description: 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.
solution: import java.io.BufferedReader;
  import java.util.StringTokenizer;
  import java.io.InputStreamReader;
  import java.io.IOException;
  import java.io.PrintWriter;
  import java.util.*;
  public class C {
  
  
     
  
      public static void main(String[] args) {
          FastReader fr = new FastReader();
          PrintWriter out = new PrintWriter(System.out, true);
          int cases = fr.nextInt();
          for(int c = 0; c < cases; c++) {
              int size = fr.nextInt();
              String num = fr.next();
              int first_zero = -1;
              for(int i = 0; i < num.length(); i++) {
                  if(num.charAt(i) == '0') {
                      first_zero = i;
                      break;
                  }
              }
  
              if(first_zero == -1) {
                  if(num.length() % 2 == 0) {out.write(writeNums(1, num.length()/2, num.length()/2+1, num.length()));}
                  else { out.write(writeNums(1, num.length()/2, num.length()/2+2, num.length())); }
              }
              else if(first_zero < num.length()/2) {
                  out.write(writeNums(first_zero+1, num.length(), first_zero+2, num.length()));
              } else {
                  out.write(writeNums(1, first_zero+1, 1, first_zero));
              }
  
  
             
              
          }
          out.close();
      }
  
      static String writeNums(int a, int b, int c, int d) {
          return a + " " + b + " " + c + " " + d + "\n";
      }
  
  
      static class FastReader {
  
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              this.br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() { 
              while (st == null || !st.hasMoreElements()) {
                  try{
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() { return Integer.parseInt(next()); }
  
          long nextLong() { return Long.parseLong(next()); }
  
          double nextDouble() { return Double.parseDouble(next()); }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
      }
  }
  