There are $n$ districts in the town, the $i$-th district belongs to the $a_i$-th bandit gang. Initially, no districts are connected to each other.

You are the mayor of the city and want to build $n-1$ two-way roads to connect all districts (two districts can be connected directly or through other connected districts).

If two districts belonging to the same gang are connected directly with a road, this gang will revolt.

You don't want this so your task is to build $n-1$ two-way roads in such a way that all districts are reachable from each other (possibly, using intermediate districts) and each pair of directly connected districts belong to different gangs, or determine that it is impossible to build $n-1$ roads to satisfy all the conditions.

You have to answer $t$ independent test cases.


-----Input-----

The first line of the input contains one integer $t$ ($1 \le t \le 500$) — the number of test cases. Then $t$ test cases follow.

The first line of the test case contains one integer $n$ ($2 \le n \le 5000$) — the number of districts. The second line of the test case contains $n$ integers $a_1, a_2, \ldots, a_n$ ($1 \le a_i \le 10^9$), where $a_i$ is the gang the $i$-th district belongs to.

It is guaranteed that the sum of $n$ does not exceed $5000$ ($\sum n \le 5000$).


-----Output-----

For each test case, print:

  NO on the only line if it is impossible to connect all districts satisfying the conditions from the problem statement.  YES on the first line and $n-1$ roads on the next $n-1$ lines. Each road should be presented as a pair of integers $x_i$ and $y_i$ ($1 \le x_i, y_i \le n; x_i \ne y_i$), where $x_i$ and $y_i$ are two districts the $i$-th road connects. 

For each road $i$, the condition $a[x_i] \ne a[y_i]$ should be satisfied. Also, all districts should be reachable from each other (possibly, using intermediate districts).


-----Example-----
Input
4
5
1 2 2 1 3
3
1 1 1
4
1 1000 101 1000
4
1 2 3 4

Output
YES
1 3
3 5
5 4
1 2
NO
YES
1 2
2 3
3 4
YES
1 2
1 3
1 4
[{"code": "'''\n    Auther: ghoshashis545 Ashis Ghosh\n    College: jalpaiguri Govt Enggineering College\n\n'''\nfrom os import path\nimport sys\nfrom heapq import heappush,heappop\nfrom functools import cmp_to_key as ctk\nfrom collections import deque,defaultdict as dd \nfrom bisect import bisect,bisect_left,bisect_right,insort,insort_left,insort_right\nfrom itertools import permutations\nfrom datetime import datetime\nfrom math import ceil,sqrt,log,gcd\ndef ii():return int(input())\ndef si():return input().rstrip()\ndef mi():return list(map(int,input().split()))\ndef li():return list(mi())\nabc='abcdefghijklmnopqrstuvwxyz'\nmod=1000000007\n# mod=998244353\ninf = float(\"inf\")\nvow=['a','e','i','o','u']\ndx,dy=[-1,1,0,0],[0,0,1,-1]\n\ndef bo(i):\n    return ord(i)-ord('a')\n\nfile = 1\n\n\n\n    \ndef solve():\n\n\n    \n\n    for _ in range(ii()):\n\n        n = ii()\n        a = li()\n        if len(set(a)) == 1:\n            print('NO')\n            continue\n        print('YES')\n        ans,p = [],[]\n        x = a[0]\n        y = -1\n        for i in range(1,n):\n            if a[i] != x:\n                y = i+1\n                ans.append([1,i+1])\n            else:\n                p.append(i+1)\n\n        for i in p:\n            ans.append([y,i])\n\n        for i in ans:\n            print(*i)\n\n        \n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n        \ndef __starting_point():\n\n    if(file):\n\n        if path.exists('input.txt'):\n            sys.stdin=open('input.txt', 'r')\n            sys.stdout=open('output.txt','w')\n        else:\n            input=sys.stdin.readline\n    solve()\n\n__starting_point()", "passed": true, "time": 0.15, "memory": 16812.0, "status": "done"}, {"code": "import sys\ninput=sys.stdin.readline\nT=int(input())\nfor _ in range(T):\n    n=int(input())\n    A=list(map(int,input().split()))\n    D={}\n    flag=0\n    for i in range(n):\n        if A[i] in D:\n            D[A[i]].append(i+1)\n\n        else:\n            D[A[i]]=[i+1]\n            flag=flag+1\n\n    if (flag==1):\n        print(\"NO\")\n\n    else:\n        print(\"YES\")\n\n        v=A[0]\n        v1=0\n        for i in range(n):\n            if (i+1 not in D[v]):\n                print(D[v][0],i+1)\n                v1=i+1\n\n        l=len(D[v])\n        for i in range(1,l):\n            print(v1,D[v][i])\n", "passed": true, "time": 0.14, "memory": 14244.0, "status": "done"}, {"code": "\"\"\"\n                            pppppppppppppppppppp\n                         ppppp  ppppppppppppppppppp\n                      ppppppp    ppppppppppppppppppppp\n                      pppppppp  pppppppppppppppppppppp\n                      pppppppppppppppppppppppppppppppp\n                              pppppppppppppppppppppppp\n       ppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppp\n      pppppppppppppppppppppppppppppppppppppppppppppppp  ppppppppppppppppppppp\n     ppppppppppppppppppppppppppppppppppppppppppppppppp  pppppppppppppppppppppp\n    ppppppppppppppppppppppppppppppppppppppppppppppp    pppppppppppppppppppppppp\n   pppppppppppppppppppppppppppppppppppppppppppppp     pppppppppppppppppppppppppp\n  ppppppppppppppppppppppppppppppppppppppppppppp      pppppppppppppppppppppppppppp\n  pppppppppppppppppppppppppppppppp               pppppppppppppppppppppppppppppppp\n  pppppppppppppppppppppppppppp     pppppppppppppppppppppppppppppppppppppppppppppp\n  ppppppppppppppppppppppppppp    pppppppppppppppppppppppppppppppppppppppppppppppp\n    pppppppppppppppppppppppp  pppppppppppppppppppppppppppppppppppppppppppppppppp\n     ppppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppppp\n      pppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppppp\n       ppppppppppppppppppppp  ppppppppppppppppppppppppppppppppppppppppppppp\n                              pppppppppppppppppppppppp\n                              pppppppppppppppppppppppppppppppp\n                              pppppppppppppppppppppp  pppppppp\n                              ppppppppppppppppppppp    ppppppp\n                                 ppppppppppppppppppp  ppppp\n                                    pppppppppppppppppppp\n\"\"\"\n\n\nimport sys\nfrom functools import lru_cache, cmp_to_key\nfrom heapq import merge, heapify, heappop, heappush, nsmallest\nfrom math import ceil, floor, gcd, fabs, factorial, fmod, sqrt, inf, log\nfrom collections import defaultdict as dd, deque, Counter as C\nfrom itertools import combinations as comb, permutations as perm\nfrom bisect import bisect_left as bl, bisect_right as br, bisect\nfrom time import perf_counter\nfrom fractions import Fraction\nfrom decimal import Decimal\nfrom copy import deepcopy\n# sys.setrecursionlimit(pow(10, 6))\n# sys.stdin = open(\"input.txt\", \"r\")\n# sys.stdout = open(\"output.txt\", \"w\")\n# mod = 10 ** 9 + 7\nmod2 = 998244353\ndef data(): return sys.stdin.readline().strip()\ndef out(var, end=\"\\n\"): sys.stdout.write(str(var)+end)\ndef outa(*var, end=\"\\n\"): sys.stdout.write(' '.join(map(str, var)) + end)\ndef l(): return list(sp())\ndef sl(): return list(ssp())\ndef sp(): return list(map(int, data().split()))\ndef ssp(): return list(map(str, data().split()))\ndef l1d(n, val=0): return [val for i in range(n)]\ndef l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]\n\n\nfor _ in range(int(data())):\n    n = int(data())\n    arr = l()\n    if len(set(arr)) == 1:\n        out(\"NO\")\n        continue\n    out(\"YES\")\n    no, L = -1, []\n    for i in range(1, n):\n        if arr[i] == arr[0]:\n            L.append(i + 1)\n        else:\n            no = i + 1\n            outa(1, i + 1)\n    for i in L:\n        outa(no, i)\n", "passed": true, "time": 0.15, "memory": 14220.0, "status": "done"}]