Rings
 Frodo was caught by Saruman. He tore a pouch from Frodo's neck, shook out its contents —there was a pile of different rings: gold and silver..."How am I to tell which is the One?!" the mage howled."Throw them one by one into the Cracks of Doom and watch when Mordor falls!" Somewhere in a parallel Middle-earth, when Saruman caught Frodo, he only found $$$n$$$ rings. And the $$$i$$$-th ring was either gold or silver. For convenience Saruman wrote down a binary string $$$s$$$ of $$$n$$$ characters, where the $$$i$$$-th character was 0 if the $$$i$$$-th ring was gold, and 1 if it was silver.Saruman has a magic function $$$f$$$, which takes a binary string and returns a number obtained by converting the string into a binary number and then converting the binary number into a decimal number. For example, $$$f(001010) = 10, f(111) = 7, f(11011101) = 221$$$.Saruman, however, thinks that the order of the rings plays some important role. He wants to find $$$2$$$ pairs of integers $$$(l_1, r_1), (l_2, r_2)$$$, such that: $$$1 \le l_1 \le n$$$, $$$1 \le r_1 \le n$$$, $$$r_1-l_1+1\ge \lfloor \frac{n}{2} \rfloor$$$  $$$1 \le l_2 \le n$$$, $$$1 \le r_2 \le n$$$, $$$r_2-l_2+1\ge \lfloor \frac{n}{2} \rfloor$$$  Pairs $$$(l_1, r_1)$$$ and $$$(l_2, r_2)$$$ are distinct. That is, at least one of $$$l_1 \neq l_2$$$ and $$$r_1 \neq r_2$$$ must hold. Let $$$t$$$ be the substring $$$s[l_1:r_1]$$$ of $$$s$$$, and $$$w$$$ be the substring $$$s[l_2:r_2]$$$ of $$$s$$$. Then there exists non-negative integer $$$k$$$, such that $$$f(t) = f(w) \cdot k$$$.Here substring $$$s[l:r]$$$ denotes $$$s_ls_{l+1}\ldots s_{r-1}s_r$$$, and $$$\lfloor x \rfloor$$$ denotes rounding the number down to the nearest integer.Help Saruman solve this problem! It is guaranteed that under the constraints of the problem at least one solution exists.
In the first testcase $$$f(t) = f(1111) = 15$$$, $$$f(w) = f(101) = 5$$$.In the second testcase $$$f(t) = f(111000111) = 455$$$, $$$f(w) = f(000111) = 7$$$.In the third testcase $$$f(t) = f(0000) = 0$$$, $$$f(w) = f(1000) = 8$$$.In the fourth testcase $$$f(t) = f(11011) = 27$$$, $$$f(w) = f(011) = 3$$$.In the fifth testcase $$$f(t) = f(001111) = 15$$$, $$$f(w) = f(011) = 3$$$.
Each test contains multiple test cases.The first line contains one positive integer $$$t$$$ ($$$1 \le t \le 10^3$$$), denoting the number of test cases. Description of the test cases follows.The first line of each test case contains one positive integer $$$n$$$ ($$$2 \le n \le 2 \cdot 10^4$$$) — length of the string.The second line of each test case contains a non-empty binary string of length $$$n$$$.It is guaranteed that the sum of $$$n$$$ over all test cases does not exceed $$$10^5$$$.
For every test case print four integers $$$l_1$$$, $$$r_1$$$, $$$l_2$$$, $$$r_2$$$, which denote the beginning of the first substring, the end of the first substring, the beginning of the second substring, and the end of the second substring, respectively.If there are multiple solutions, print any.
import sys
input = sys.stdin.readline

def readList():
    return list(map(int, input().split()))

def readInt():
    return int(input())

def readInts():
    return map(int, input().split())

def readStr():
    return input().strip()

def solve():
    (n, s) = (readInt(), readStr())
    for i in range(n - 1, -1, -1):
        if s[i] == '0':
            if i >= n // 2:
                print(1, i + 1, 1, i)
            else:
                print(i + 1, n, i + 2, n)
            return
    print(1, n - 1, 2, n)
    return
for _ in range(readInt()):
    solve()
for _ in range(int(input())):
    n = int(input())
    s = input()
    if '0' in s:
        i = s.index('0')
        if i < n // 2:
            print(i + 1, n, i + 2, n)
        else:
            print(1, i + 1, 1, i)
    else:
        print(1, n // 2, n // 2 + 1, n // 2 + n // 2)
import sys
import math
import heapq
import collections
input = sys.stdin.readline
testcases = int(input())

def solution():
    n = int(input())
    rings = input().rstrip()
    window = n // 2
    for i in range(n):
        if rings[i] == '0':
            if i < window:
                return (i + 2, n, i + 1, n)
            else:
                return (1, i + 1, 1, i)
    return (1, 2 * window, 1, window)
for i in range(testcases):
    print(' '.join(map(str, solution())))