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 math import *
   
  #from math import factorial as fact, comb as ncr 
  from bisect import bisect_left as bl
  from bisect import bisect_right as br
  from collections import Counter as ctr
  from collections import deque as dq
  
  from array import array
  from re import search
   
  li=lambda : list(map(int,input().split()))
  arr=lambda a: array('i',a)
  bi=lambda n: bin(n).replace("0b", "")
  yn=lambda f: print('NYOE S'[f::2])
  sbstr=lambda a,s: search('.*'.join(a),s)
  
  '''def isprime(a):
      if a==2: return True
      elif a%2==0: return False
      n=3
      while n<=a**0.5:
          if a%n==0:
              return False
          n+=2
      return True
  '''
  '''def nf(n,a):
      b=a*a
      c=b*a
      d=c*a
      e=n
      ct=0
      while e%a==0:
          if e%d==0:
              ct+=4
              e=e//d
          if e%c==0:
              ct+=3
              e=e//c
          if e%b==0:
              ct+=2
              e=e//b
          if e%a==0:
              ct+=1
              e=e//a
      return ct
  '''
  
  def solve():
      for _ in range(int(input())):
          n=int(input())
          a=li()[::-1]
          c=0
          m=max(a)
          t=a[0]
          for i in range(n):
              if a[i]>t:
                  t=a[i]
                  c+=1
              elif t==m:
                  break
          print(c)
  
  
  solve()