https://codeforces.com/contest/1562/problem/C
#include<iostream>
#include<bits/stdc++.h>
using namespace std;

const int N = 1e5 + 10;


void solve() {
    string s;
    int n;
    cin >> n;
    cin >> s;
    int k = - 1;
    for (int i = 0; i < n; i ++) {
        if (s[i] == '0') {
            k = 1;
            if (i>= n / 2) cout << 1 << ' ' << i + 1 << " " << 1 << " " << i << endl;
            else cout << i + 2 << " " <<n << " " <<  i + 1  << ' ' <<  n << endl;
            break;
        }
        
    }
    if (k == - 1) {
        cout << 1 << ' ' << n- 1 << ' ' << 2 << " " << n << endl;
    }
}
int main () {
    int t;
     cin >> t;
     while (t --) solve();   
    return 0;
}

考虑哪些k容易计算，1, 2 0 先考虑极端形况[1, n - 1] [2, n] 只要找到一个0位置为k，如果这个k在左半边，那么[k, n] [k + 1, n] 否则[1, k] [1, k - 1];