human
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...
assistant
#Here is Python code to find 2 pairs of integers (l_1, r_1), (l_2, r_2), such that:
.

import math
from collections import defaultdict
from heapq import heappush, heappop
 
DEBUG = True
 
def log(*args, **kwargs):
    if DEBUG:
        print(*args, **kwargs)
 
def ri():
    return int(input())
 
def rl(f=int):
    return list(map(f, input().split()))
 
def rs():
    return input()
 
class Solution:
 
    def __init__(self):
        pass
    
    def run(self):
 
        n = ri()
        s = rs()
 
        zeros = []
 
        for i in range(len(s)):
            if s[i] == '0':
                zeros.append(i)
 
        
        if len(zeros) == 0:
            if n % 2 == 0:
                print(1, n // 2, n // 2 + 1, n)
            else:
                print(1, n // 2 + 1, n // 2 + 1, n)
            return
 
        else:
            z = zeros[0]
            if z >= n // 2:
                print(1, z + 1, 1, z)
            else:
                print(z + 2, n, z + 1, n)
            return
            
        
            
 
        
 
        
        
 
        
 
        
        
        
            
                
if __name__ == '__main__':
    
    t = int(input())
    s = Solution()
    #print(s.run())
    #s.run()
    for i in range(t):
        s.run()
    #    print(s.run())