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 bisect
  import heapq
  import math
  import collections
  import sys
  import copy
  from functools import reduce
  import decimal
  from io import BytesIO, IOBase
  import os
  import itertools
  import functools
  from types import GeneratorType
  
  sys.setrecursionlimit(10 ** 9)
  decimal.getcontext().rounding = decimal.ROUND_HALF_UP
  
  graphDict = collections.defaultdict
  
  queue = collections.deque
  
  
  ################## pypy deep recursion handling ##############
  # Author = @pajenegod
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          to = f(*args, **kwargs)
          if stack:
              return to
          else:
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          return to
                      to = stack[-1].send(to)
  
      return wrappedfunc
  
  
  ################## Graphs ###################
  class Graphs:
      def __init__(self):
          self.graph = graphDict(set)
  
      def add_edge(self, u, v):
          self.graph[u].add(v)
          self.graph[v].add(u)
  
      def dfs_utility(self, nodes, visited_nodes, psf):
          global ans
          if len(visited_nodes.keys()) == n:
              ans = psf
              return
          visited_nodes[nodes] = None
          for neighbour in self.graph[nodes]:
              if neighbour not in visited_nodes:
                  self.dfs_utility(neighbour, visited_nodes, psf + [neighbour])
          visited_nodes.pop(nodes)
          return
  
      def dfs(self, node):
          global ans
          Visited = collections.OrderedDict()
          self.dfs_utility(node, Visited, [node])
  
      def bfs(self, node):
          visited = set()
          if node not in visited:
              queue.append(node)
              visited.add(node)
          while queue:
              parent = queue.popleft()
              print(parent)
              for item in self.graph[parent]:
                  if item not in visited:
                      queue.append(item)
                      visited.add(item)
  
  
  ################### Tree Implementaion ##############
  class Tree:
      def __init__(self, data):
          self.data = data
          self.left = None
          self.right = None
  
  
  def inorder(node, lis):
      if node:
          inorder(node.left, lis)
          lis.append(node.data)
          inorder(node.right, lis)
      return lis
  
  
  def leaf_node_sum(root):
      if root is None:
          return 0
      if root.left is None and root.right is None:
          return root.data
      return leaf_node_sum(root.left) + leaf_node_sum(root.right)
  
  
  def hight(root):
      if root is None:
          return -1
      if root.left is None and root.right is None:
          return 0
      return max(hight(root.left), hight(root.right)) + 1
  
  
  #################################################
  
  def rounding(n):
      return int(decimal.Decimal(f'{n}').to_integral_value())
  
  
  def factors(n):
      return set(reduce(list.__add__,
                        ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0), []))
  
  
  def p_sum(array):
      return list(itertools.accumulate(array))
  
  
  ################################ <fast I/O> ###########################################
  BUFSIZE = 8192
  
  
  class FastIO(IOBase):
      newlines = 0
  
      def __init__(self, file):
          self._fd = file.fileno()
          self.buffer = BytesIO()
          self.writable = "x" in file.mode or "r" not in file.mode
          self.write = self.buffer.write if self.writable else None
  
      def read(self):
          while True:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              if not b:
                  break
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines = 0
          return self.buffer.read()
  
      def readline(self, **kwargs):
          while self.newlines == 0:
              b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
              self.newlines = b.count(b"\n") + (not b)
              ptr = self.buffer.tell()
              self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
          self.newlines -= 1
          return self.buffer.readline()
  
      def flush(self):
          if self.writable:
              os.write(self._fd, self.buffer.getvalue())
              self.buffer.truncate(0), self.buffer.seek(0)
  
  
  class IOWrapper(IOBase):
      def __init__(self, file):
          self.buffer = FastIO(file)
          self.flush = self.buffer.flush
          self.writable = self.buffer.writable
          self.write = lambda s: self.buffer.write(s.encode("ascii"))
          self.read = lambda: self.buffer.read().decode("ascii")
          self.readline = lambda: self.buffer.readline().decode("ascii")
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  
  #############################################<I/O Region >##############################################
  
  
  def inp():
      return sys.stdin.readline().strip()
  
  
  def map_inp(v_type):
      return map(v_type, inp().split())
  
  
  def list_inp(v_type):
      return list(map_inp(v_type))
  
  
  ######################################## Solution ####################################
  
  for _ in range(int(inp())):
      n = int(inp())
      s = inp()
      if s.count("0"):
          x = s.index("0")
          if n - (x + 2) + 1 >= n // 2:
              print(x + 1, n, x + 2, n)
          else:
              x = s.rfind("0")
              print(1, x + 1, 1, x)
      else:
          print(1, n-1, 2, n)
  