description: You are given an array a of length n.
  
  Let's define the eversion operation. Let x = a_n. Then array a is partitioned into two parts: left and right. The left part contains the elements of a that are not greater than x (≤ x). The right part contains the elements of a that are strictly greater than x (> x). The order of elements in each part is kept the same as before the operation, i. e. the partition is stable. Then the array is replaced with the concatenation of the left and the right parts.
  
  For example, if the array a is [2, 4, 1, 5, 3], the eversion goes like this: [2, 4, 1, 5, 3] → [2, 1, 3], [4, 5] → [2, 1, 3, 4, 5].
  
  We start with the array a and perform eversions on this array. We can prove that after several eversions the array a stops changing. Output the minimum number k such that the array stops changing after k eversions.
  
  Input
  
  Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.
  
  The first line contains a single integer n (1 ≤ n ≤ 2 ⋅ 10^5).
  
  The second line contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ 10^9).
  
  It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.
  
  Output
  
  For each test case print a single integer k — the number of eversions after which the array stops changing.
  
  Example
  
  Input
  
  
  3
  5
  2 4 1 5 3
  5
  5 3 2 4 1
  4
  1 1 1 1
  
  
  Output
  
  
  1
  2
  0
  
  Note
  
  Consider the fist example.
  
    * The first eversion: a = [1, 4, 2, 5, 3], x = 3. [2, 4, 1, 5, 3] → [2, 1, 3], [4, 5] → [2, 1, 3, 4, 5]. 
    * The second and following eversions: a = [2, 1, 3, 4, 5], x = 5. [2, 1, 3, 4, 5] → [2, 1, 3, 4, 5], [] → [2, 1, 3, 4, 5]. This eversion does not change the array, so the answer is 1. 
  
  
  
  Consider the second example. 
  
    * The first eversion: a = [5, 3, 2, 4, 1], x = 1. [5, 3, 2, 4, 1] → [1], [5, 3, 2, 4] → [1, 5, 3, 2, 4]. 
    * The second eversion: a = [1, 5, 3, 2, 4], x = 4. [1, 5, 3, 2, 4] → [1, 3, 2, 4], [5] → [1, 3, 2, 4, 5]. 
    * The third and following eversions: a = [1, 3, 2, 4, 5], x = 5. [1, 3, 2, 4, 5] → [1, 3, 2, 4, 5], [] → [1, 3, 2, 4, 5]. This eversion does not change the array, so the answer is 2. 
solution: from __future__ import division, print_function
  
  import os,sys
  from io import BytesIO, IOBase
  
  if sys.version_info[0] < 3:
      from __builtin__ import xrange as range
      from future_builtins import ascii, filter, hex, map, oct, zip
  
  from bisect import bisect_left as lower_bound, bisect_right as upper_bound 
  def so():  return int(input())
  def st():  return input()
  def mj():  return map(int,input().strip().split(" "))
  def msj(): return map(str,input().strip().split(" "))
  def le():  return list(map(int,input().split()))
  def rc():  return map(float,input().split())
  def lebe():return list(map(int, input()))
  
  def dmain():
      sys.setrecursionlimit(1000000)
      threading.stack_size(1024000)
      thread = threading.Thread(target=main)
      thread.start()
  def joro(L):
      return(''.join(map(str, L)))
  
  
  def decimalToBinary(n): return bin(n).replace("0b","")
  
  
  def isprime(n):
      for i in range(2,int(n**0.5)+1):
          if n%i==0:
              return False
      return True
  def npr(n, r):
      return factorial(n) // factorial(n - r) if n >= r else 0
   
   
  def ncr(n, r):
      return factorial(n) // (factorial(r) * factorial(n - r)) if n >= r else 0
   
   
  def lower_bound(li, num):
      answer = -1
      start = 0
      end = len(li) - 1
   
      while (start <= end):
          middle = (end + start) // 2
          if li[middle] >= num:
              answer = middle
              end = middle - 1
          else:
              start = middle + 1
      return answer  # min index where x is not less than num
   
   
  def upper_bound(li, num):
      answer = -1
      start = 0
      end = len(li) - 1
   
      while (start <= end):
          middle = (end + start) // 2
   
          if li[middle] <= num:
              answer = middle
              start = middle + 1
   
          else:
              end = middle - 1
      return answer  # max index where x is not greater than num
  def tir(a,b,c):
      if(0==c):
          return 1
      if(len(a)<=b):
          return 0
      
      if(c!=-1):
          return (tir(a,1+b,c+a[b]) or tir(a,b+1,c-a[b]) or tir(a,1+b,c)) 
          
          
      else:
          return (tir(a,1+b,a[b]) or tir(a,b+1,-a[b]) or tir(a,1+b,-1))
  hoi=int(2**20)        
   
  def abs(x):
      return x if x >= 0 else -x
   
   
  def binary_search(li, val, lb, ub):
      # print(lb, ub, li)
      ans = -1
      while (lb <= ub):
          mid = (lb + ub) // 2
          # print('mid is',mid, li[mid])
          if li[mid] > val:
              ub = mid - 1
          elif val > li[mid]:
              lb = mid + 1
          else:
              ans = mid  # return index
              break
      return ans
   
   
  def kadane(x):  # maximum sum contiguous subarray
      sum_so_far = 0
      current_sum = 0
      for i in x:
          current_sum += i
          if current_sum < 0:
              current_sum = 0
          else:
              sum_so_far = max(sum_so_far, current_sum)
      return sum_so_far
   
   
  def pref(li):
      pref_sum = [0]
      for i in li:
          pref_sum.append(pref_sum[-1] + i)
      return pref_sum
   
   
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n + 1)]
      p = 2
      li = []
      while (p * p <= n):
          if (prime[p] == True):
              for i in range(p * p, n + 1, p):
                  prime[i] = False
          p += 1
   
      for p in range(2, len(prime)):
          if prime[p]:
              li.append(p)
      return li
   
   
  def primefactors(n):
      factors = []
      while (n % 2 == 0):
          factors.append(2)
          n //= 2
      for i in range(3, int(sqrt(n)) + 1, 2):  # only odd factors left
          while n % i == 0:
              factors.append(i)
              n //= i
      if n > 2:  # incase of prime
          factors.append(n)
      return factors
           
      
  def read():
      sys.stdin  = open('input.txt', 'r')  
      sys.stdout = open('output.txt', 'w') 
  def tr(n):
      return n*(n+1)//2
  boi=int(1e9-7)
  doi=int(1e9+7)
  
          
      
  def iu():
      import sys
      import math as my
      input=sys.stdin.readline
      m=so()
      L=le()
      bec=0
      z=L[m-1]
      for i in range(m-2,-1,-1):
          if(L[i]>z):
              z=L[i]
              bec+=1
      print(bec)
      
      
      
  def main():
      for i in range(so()):
          iu()
          
      
      
      
      
      
      
      
      
      
      
  
      
  
      
          
          
              
      
      
                  
              
                  
                  
                  
                  
              
      
              
              
      
      
      
      
          
              
      
             
              
              
          
      
          
      
      
          
          
          
  
              
                      
                  
                      
                      
                      
                      
                      
                      
          
                  
          
          
         
      
             
            
            
                  
              
          
                  
      
          
          
          
         
              
                  
          
  
  
  
  
  # region fastio
  # template taken from https://github.com/cheran-senthil/PyRival/blob/master/templates/template.py
  
  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):
          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")
  
  
  def print(*args, **kwargs):
      """Prints the values to a stream, or to sys.stdout by default."""
      sep, file = kwargs.pop("sep", " "), kwargs.pop("file", sys.stdout)
      at_start = True
      for x in args:
          if not at_start:
              file.write(sep)
          file.write(str(x))
          at_start = False
      file.write(kwargs.pop("end", "\n"))
      if kwargs.pop("flush", False):
          file.flush()
  
  
  if sys.version_info[0] < 3:
      sys.stdin, sys.stdout = FastIO(sys.stdin), FastIO(sys.stdout)
  else:
      sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  
  input = lambda: sys.stdin.readline().rstrip("\r\n")
  
  # endregion
  
  
  if __name__ == "__main__":
      #read()
      main()
      #dmain()
  
  # Comment Read()