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: from math import gcd
  from bisect import bisect_left, bisect_right
  from itertools import combinations
  from itertools import permutations
  from bisect import bisect_left
  from math import ceil
  from heapq import heapify, heappush, heappop, nsmallest
  import bisect
  from math import pi
  from collections import deque
  from math import factorial
  from math import log, ceil
  from collections import defaultdict
  from math import *
  from sys import stdin, stdout
  import itertools
  import os
  import sys
  import threading
  from collections import deque, Counter, OrderedDict, defaultdict
  from heapq import *
  # from math import ceil, floor, log, sqrt, factorial, pow, pi, gcd
  # from bisect import bisect_left,bisect_right
  # from decimal import *,threading
  from fractions import Fraction
  mod = int(pow(10, 9)+7)
  # mod = 998244353
  
  
  def ii(): return int(input())
  
  
  def si(): return str(input())
  
  
  def mi(): return map(int, input().split())
  
  
  def li1(): return list(mi())
  
  
  def fii(): return int(stdin.readline())
  
  
  def fsi(): return str(stdin.readline())
  
  
  def fmi(): return map(int, stdin.readline().split())
  
  
  def fli(): return list(fmi())
  
  
  abd = {'a': 0, 'b': 1, 'c': 2, 'd': 3, 'e': 4, 'f': 5, 'g': 6, 'h': 7, 'i': 8, 'j': 9, 'k': 10, 'l': 11, 'm': 12,
      'n': 13, 'o': 14, 'p': 15, 'q': 16, 'r': 17, 's': 18, 't': 19, 'u': 20, 'v': 21, 'w': 22, 'x': 23, 'y': 24,
      'z': 25}
  
  
  def getKey(item): return item[0]
  
  
  def sort2(l): return sorted(l, key=getKey)
  
  
  def d2(n, m, num): return [[num for x in range(m)] for y in range(n)]
  
  
  def isPowerOfTwo(x): return (x and (not (x & (x - 1))))
  
  
  def decimalToBinary(n): return bin(n).replace("0b", "")
  
  
  def ntl(n): return [int(i) for i in str(n)]
  
  
  def powerMod(x, y, p):
      res = 1
      x %= p
      while y > 0:
          if y & 1:
              res = (res * x) % p
          y = y >> 1
          x = (x * x) % p
      return res
  
  
  graph = defaultdict(list)
  visited = [0] * 1000000
  col = [-1] * 1000000
  
  
  def bfs(d, v):
      q = []
      q.append(v)
      visited[v] = 1
      while len(q) != 0:
          x = q[0]
          q.pop(0)
          for i in d[x]:
              if visited[i] != 1:
                  visited[i] = 1
                  q.append(i)
          print(x)
  
  
  def make_graph(e):
      d = {}
      for i in range(e):
          x, y = mi()
          if x not in d:
              d[x] = [y]
          else:
              d[x].append(y)
          if y not in d:
              d[y] = [x]
          else:
              d[y].append(x)
      return d
  
  
  def gr2(n):
      d = defaultdict(list)
      for i in range(n):
          x, y = mi()
          d[x].append(y)
      return d
  
  
  def connected_components(graph):
      seen = set()
  
      def dfs(v):
          vs = set([v])
          component = []
          while vs:
              v = vs.pop()
              seen.add(v)
              vs |= set(graph[v]) - seen
              component.append(v)
          return component
  
      ans = []
      for v in graph:
          if v not in seen:
              d = dfs(v)
              ans.append(d)
      return ans
  
  
  def primeFactors(n):
      s = set()
      while n % 2 == 0:
          s.add(2)
          n = n // 2
      for i in range(3, int(sqrt(n)) + 1, 2):
          while n % i == 0 and i % 2 == 1:
              s.add(i)
              n = n // i
      if n > 2 and n % 2 == 1:
          s.add(n)
      return s
  
  
  def find_all(a_str, sub):
      start = 0
      while True:
          start = a_str.find(sub, start)
          if start == -1:
              return
          yield start
          start += len(sub)
  
  
  def SieveOfEratosthenes(n, isPrime):
      isPrime[0] = isPrime[1] = False
      for i in range(2, n):
          isPrime[i] = True
      p = 2
      while (p * p <= n):
          if (isPrime[p] == True):
              i = p * p
              while (i <= n):
                  isPrime[i] = False
                  i += p
          p += 1
      return isPrime
  
  
  def dijkstra(edges, f, t):
      g = defaultdict(list)
      for l, r, c in edges:
          g[l].append((c, r))
  
      q, seen, mins = [(0, f, ())], set(), {f: 0}
      while q:
          (cost, v1, path) = heappop(q)
          if v1 not in seen:
              seen.add(v1)
              path = (v1, path)
              if v1 == t:
                  return (cost, path)
  
              for c, v2 in g.get(v1, ()):
                  if v2 in seen:
                      continue
                  prev = mins.get(v2, None)
                  next = cost + c
                  if prev is None or next < prev:
                      mins[v2] = next
                      heappush(q, (next, v2, path))
      return float("inf")
  
  
  def binsearch(a, l, r, x):
      while l <= r:
          mid = l + (r-1)//2
          if a[mid]:
              return mid
          elif a[mid] > x:
              l = mid-1
          else:
              r = mid+1
      return -1
  
  
  # def input():
  #    return stdin.buffer.readline()
  
  
  def readTree(n):
      adj = [set() for _ in range(n)]
      for _ in range(n-1):
          u, v = map(int, input().split())
          adj[u-1].add(v-1)
          adj[v-1].add(u-1)
      return adj
  
  
  def treeOrderByDepth(n, adj, root=0):
      parent = [-2] + [-1]*(n-1)
      ordered = []
      q = deque()
      q.append(root)
      depth = [0] * n
      while q:
          c = q.popleft()
          ordered.append(c)
          for a in adj[c]:
              if parent[a] == -1:
                  parent[a] = c
                  depth[a] = depth[c] + 1
                  q.append(a)
      return (ordered, parent, depth)
  
  
  def isSubSequence(str1, str2):
      m = len(str1)
      n = len(str2)
      j = 0
      i = 0
      while j < m and i < n:
          if str1[j] == str2[i]:
              j = j+1
          i = i + 1
      return j == m
  
  
  def nextPowerOf2(n):
      count = 0
      if (n and not(n & (n - 1))):
          return n
      while(n != 0):
          n >>= 1
          count += 1
      return 1 << count
  
  
  def lastPowerOf2(n):
      count = -1
      if (n and not(n & (n - 1))):
          return n
      while(n != 0):
          n >>= 1
          count += 1
      return 1 << count
  
  
  def cou(n):
      c = 0
      while n > 1:
          c += 1
          n //= 2
      return c
  
  
  def sortsec(l):
      return sorted(l, key=lambda x: x[1])
  
  
  def BinarySearch(a, x):
      i = bisect_left(a, x)
      if i:
          return (i-1)
      else:
          return - 1
  
  
  def subarray(A):
      r = set()
  
      p = {0}
  
      for x in A:
          p = {x | y for y in p} | {x}
          r |= p
  
      return len(r)
  
  
  def setBitNumber(n):
      n |= n >> 1
      n |= n >> 2
      n |= n >> 4
      n |= n >> 8
      n |= n >> 16
      n = n + 1
      return (n >> 1)
  
  
  def countSetBits(n):
      count = 0
      while (n):
          count += n & 1
          n >>= 1
      return count
  
  
  def allsub(l):
      n = len(l)
      a = []
      for i in range(n - 1):
          a.append([l[:i], l[i:]])
          a.append([l[i:], l[:i]])
      return a
  
  
  def closestMultiple(n, x):
      if x > n:
          return x
      z = (int)(x / 2)
      n = n + z
      n = n - (n % x)
      return n
  
  
  def block(x):
      v = []
      res = []
      while (x > 0):
          v.append(int(x % 2))
          x = int(x / 2)
      for i in range(0, len(v)):
          if (v[i] == 1):
              res.append(i)
      return res
      
  for _ in range(ii()):
      n=ii()
      p=0
      s=si()
      for i in range(1,n+1):
          if s[i-1] == '0':
              if i<=n//2:
                  print(i,i+n//2,i+1,i+n//2)
                  p=1
                  break
              else:
                  print(i-n//2,i,i-n//2,i-1)
                  p=1
                  break
      if p==0:
          print(1,n-1,2,n)