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 ≤ l_1 ≤ n, 1 ≤ r_1 ≤ n, r_1-l_1+1≥ ⌊ n/2 ⌋ 
  * 1 ≤ l_2 ≤ n, 1 ≤ r_2 ≤ n, r_2-l_2+1≥ ⌊ n/2 ⌋ 
  * Pairs (l_1, r_1) and (l_2, r_2) are distinct. That is, at least one of l_1 ≠ l_2 and r_1 ≠ 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) ⋅ k.



Here substring s[l:r] denotes s_ls_{l+1}… s_{r-1}s_r, and ⌊ x ⌋ 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.

Input

Each test contains multiple test cases.

The first line contains one positive integer t (1 ≤ t ≤ 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 ≤ n ≤ 2 ⋅ 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.

Output

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.

Example

Input


7
6
101111
9
111000111
8
10000000
5
11011
6
001111
3
101
30
100000000000000100000000000000


Output


3 6 1 3
1 9 4 9
5 8 1 4
1 5 3 5
1 6 2 4
1 2 2 3
1 15 16 30

Note

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.
#include <bits/stdc++.h>
using namespace std;
void __print(int x) { cerr << x; }
void __print(long x) { cerr << x; }
void __print(long long x) { cerr << x; }
void __print(unsigned x) { cerr << x; }
void __print(unsigned long x) { cerr << x; }
void __print(unsigned long long x) { cerr << x; }
void __print(float x) { cerr << x; }
void __print(double x) { cerr << x; }
void __print(long double x) { cerr << x; }
void __print(char x) { cerr << '\'' << x << '\''; }
void __print(const char *x) { cerr << '\"' << x << '\"'; }
void __print(const string &x) { cerr << '\"' << x << '\"'; }
void __print(bool x) { cerr << (x ? "true" : "false"); }
template <typename T, typename V>
void __print(const pair<T, V> &x) {
  cerr << '{';
  __print(x.first);
  cerr << ',';
  __print(x.second);
  cerr << '}';
}
template <typename T>
void __print(const T &x) {
  int f = 0;
  cerr << '{';
  for (auto &i : x) cerr << (f++ ? "," : ""), __print(i);
  cerr << "}";
}
void _print() { cerr << "]\n"; }
template <typename T, typename... V>
void _print(T t, V... v) {
  __print(t);
  if (sizeof...(v)) cerr << ", ";
  _print(v...);
}
const int N = 2e5 + 1;
long long fact[N], inv[N], invfact[N];
vector<int> primes, lpf;
void sieve(int n) {
  primes.assign(1, 2);
  lpf.assign(n + 1, 2);
  for (int x = 3; x <= n; x += 2) {
    if (lpf[x] == 2) primes.push_back(lpf[x] = x);
    for (int i = 0;
         i < primes.size() && primes[i] <= lpf[x] && primes[i] * x <= n; ++i)
      lpf[primes[i] * x] = primes[i];
  }
}
long long mul(long long a, long long b, const long long MOD = 1000000007) {
  long long res = 0;
  for (a %= MOD; b > 0; a <<= 1, b >>= 1) {
    if (a >= MOD) a -= MOD;
    if (b & 1) {
      res += a;
      if (res >= MOD) res -= MOD;
    }
  }
  return res;
}
long long power(long long x, long long n, const long long MOD = 1000000007) {
  long long res = 1;
  for (; n > 0; n >>= 1) {
    if (n & 1) res = mul(res, x, MOD);
    x = mul(x, x, MOD);
  }
  return res;
}
bool mr_test(long long n, int a, long long d, int s) {
  long long x = power(a, d, n);
  if (x == 1 || x == n - 1) return false;
  for (int r = 1; r < s; ++r) {
    x = mul(x, x, n);
    if (x == n - 1) return false;
  }
  return true;
}
bool miller_rabin(long long n, int k = 5) {
  if (n < 4) return n == 2 || n == 3;
  if (n % 2 == 0 || n % 3 == 0) return false;
  int s = __builtin_ctz(n - 1);
  long long d = (n - 1) >> s;
  for (int it = 1; it <= 5; ++it) {
    int a = 2 + rand() % (n - 3);
    if (mr_test(n, a, d, s)) return false;
  }
  return true;
}
bool isPrime(long long n) {
  if (n < 2) return false;
  if (n < lpf.size()) return lpf[n] == n;
  return miller_rabin(n);
}
long long div(long long n) {
  long long ans = 1;
  for (int i = 0; i < primes.size(); i++) {
    if (primes[i] * primes[i] * primes[i] > n) break;
    long long cnt = 1;
    while (n % primes[i] == 0) {
      n /= primes[i];
      cnt++;
    }
    ans = ans * cnt;
  }
  long long x = round(sqrt(n));
  if (isPrime(n))
    ans *= 2;
  else if (x * x == n) {
    if (isPrime(x)) ans *= 3;
  } else
    ans *= 4;
  return ans;
}
int log_(int a, int b) { return (a > b - 1) ? 1 + log_(a / b, b) : 0; }
long long add(long long a, long long b, long long Mod = 1000000007) {
  if ((a += b) >= Mod)
    a -= Mod;
  else if (a < 0)
    a += Mod;
  return a;
}
long long nCr(int n, int r) {
  if (r > n) return 0;
  return mul(mul(fact[n], invfact[r]), invfact[n - r]);
}
long long gcd(long long a, long long b) { return !b ? a : gcd(b, a % b); }
void factInverse() {
  fact[0] = inv[1] = fact[1] = invfact[0] = invfact[1] = 1;
  for (long long i = 2; i < N; i++) {
    fact[i] = (fact[i - 1] * i) % 1000000007;
    inv[i] = 1000000007 - (inv[1000000007 % i] * (1000000007 / i) % 1000000007);
    invfact[i] = (inv[i] * invfact[i - 1]) % 1000000007;
  }
}
void solve(int K) {
  long long n, m, q, z, k;
  long long ans = 0, cnt = 0;
  long long x, y;
  long long l{-1}, r{-1};
  string s;
  string t;
  cin >> n >> s;
  cnt = 0;
  x = n / 2 - 1;
  ;
  for (int i = 0; i <= x; i++) {
    if (s[i] == '0') {
      cout << i + 1 << " " << n << " " << i + 2 << " " << n << "\n";
      return;
    }
  }
  for (int i = x + 1; i < n; i++) {
    if (s[i] == '0') {
      cout << 1 << " " << i + 1 << " " << 1 << " " << i << "\n";
      return;
    }
  }
  cout << 1 << " " << n << " " << 2 << " " << n << "\n";
  return;
}
int main() {
  sieve(1000009);
  long long t, k;
  k = 1;
  t = 1;
  cin >> t;
  while (t--) {
    solve(k);
    k++;
  }
  cerr << "Time elapsed: " << 1.0 * clock() / CLOCKS_PER_SEC << " s.\n";
  cout << flush;
  return 0;
}
