1288_D. Minimax Problem_5170
You are given n arrays a_1, a_2, ..., a_n; each array consists of exactly m integers. We denote the y-th element of the x-th array as a_{x, y}.

You have to choose two arrays a_i and a_j (1 ≤ i, j ≤ n, it is possible that i = j). After that, you will obtain a new array b consisting of m integers, such that for every k ∈ [1, m] b_k = max(a_{i, k}, a_{j, k}).

Your goal is to choose i and j so that the value of min _{k = 1}^{m} b_k is maximum possible.

Input

The first line contains two integers n and m (1 ≤ n ≤ 3 ⋅ 10^5, 1 ≤ m ≤ 8) — the number of arrays and the number of elements in each array, respectively.

Then n lines follow, the x-th line contains the array a_x represented by m integers a_{x, 1}, a_{x, 2}, ..., a_{x, m} (0 ≤ a_{x, y} ≤ 10^9).

Output

Print two integers i and j (1 ≤ i, j ≤ n, it is possible that i = j) — the indices of the two arrays you have to choose so that the value of min _{k = 1}^{m} b_k is maximum possible. If there are multiple answers, print any of them.

Example

Input


6 5
5 0 3 1 2
1 8 9 1 3
1 2 3 4 5
9 1 0 3 7
2 3 0 6 3
6 4 1 7 0


Output


1 5
import sys
range = xrange
input = raw_input

inp = [int(x) for x in sys.stdin.read().split()]; ii = 0

n = inp[ii]; ii += 1
m = inp[ii]; ii += 1

A = inp[ii: ii + n * m]; ii += m

a = 0
b = 10**9
while a < b:
    c = a + b + 1 >> 1

    arr = [0]*(2**m)
    for i in range(n):
        mask = 0
        for j in range(m):
            mask = (mask << 1) | +(A[i*m + j] >= c)
        arr[mask] = 1
    
    pos = False
    goal = 2**m - 1
    for i in range(goal + 1):
        if arr[i]:
            for j in range(i + 1):
                if arr[j] and i | j == goal:
                    pos = True

    if pos:
        a = c
    else:
        b = c - 1


arr = [0]*(2**m)
who = [0]*(2**m)
for i in range(n):
    mask = 0
    for j in range(m):
        mask = (mask << 1) | +(A[i*m + j] >= a)
    arr[mask] = 1
    who[mask] = i

goal = 2**m - 1
for i in range(goal + 1):
    if arr[i]:
        for j in range(i + 1):
            if arr[j] and i | j == goal:
                print who[i] + 1, who[j] + 1
                sys.exit()
