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: ### Codeforces handle - "@titanalpha"
  #Email id - shubhamnokiafan@gmail.com
  
  from collections import Counter, deque
  import math
  from random import randint
  from bisect import bisect
  import sys
  import re
  import copy
  from decimal import Decimal,getcontext
  getcontext().prec = 32
  
  MOD = 1000000007
  MAX = sys.maxsize
  MIN = -sys.maxsize
  prime = []
  
  class get:
      def int():
          return list(map(int,input().split()))
      def str():
          return input().split()
      def float():
          return list(map(float,input().split()))
      def map():
          return map(int,input().split())
  
  def sqrt(x):
      return int(math.sqrt(x))+1
  
  def gcd(*args):
      result = args[0]
      for item in args:
          result = math.gcd(result,item)
      return result
  
  def LCM(a,b):
      return (a*b)//(math.gcd(a,b))
  
  def isPrime(n):
      if n == 1:
          return False
      for i in range(2,sqrt(n)):
          if n%i == 0:
              return False
      return True
  
  def solve():
      n = int(input())
      a = get.int()
      if a[-1] == max(a):
          print(0)
          return
      max_ = MIN
      j = 0
      for i in range(n-1,-1,-1):
          if a[i] > max_:
              max_ = a[i]
              j = i
      r_max = MIN
      res = 0
      for i in range(n-1,j,-1):
          if a[i] > r_max:
              r_max = a[i]
              res += 1
      print(res)
  
  
  
  for _ in range(int(input())):
      solve()