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: import os
  import sys
  from bisect import bisect, bisect_left, insort, insort_left
  from heapq import heapify, heappop, heappush, nlargest, nsmallest
  from io import BytesIO, IOBase
  from math import inf, sqrt
  from types import GeneratorType
  
  # Fast IO for Python
  
  BUFSIZE: int = 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()
  
  
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  def input(): return sys.stdin.readline().rstrip("\r\n")
  
  
  # Bootstrap used for recursion, use yield instead of return
  
  def bootstrap(f, stack=[]):
      def wrappedfunc(*args, **kwargs):
          if stack:
              return f(*args, **kwargs)
          else:
              to = f(*args, **kwargs)
              while True:
                  if type(to) is GeneratorType:
                      stack.append(to)
                      to = next(to)
                  else:
                      stack.pop()
                      if not stack:
                          break
                      to = stack[-1].send(to)
              return to
      return wrappedfunc
  
  
  # Never use += for string concatenation
  # Always use .join for string concatenation
  # Use PyPy3 for Codeforces
  # Use Python3 for Codechef
  
  # Function to compute inverse of a under modulo m
  
  def modInverse(a, m):
  
      g = gcd(a, m)
  
      if (g != 1):
          return -1
  
      else:
          # If a and m are relatively prime,
          # then modulo inverse is a^(m-2) mode m
          return power(a, m - 2, m)
  
  
  # Function to compute x^y under modulo m
  
  def power(x, y, m):
  
      if (y == 0):
          return 1
  
      p = power(x, y // 2, m) % m
      p = (p * p) % m
  
      if(y % 2 == 0):
          return p
      else:
          return ((x * p) % m)
  
  
  # Function to return gcd of a and b
  
  def gcd(a, b):
      if (a == 0):
          return b
  
      return gcd(b % a, a)
  
  
  # Sieve of Eratosthenes to get prime upto n
  
  def SieveOfEratosthenes(n):
      prime = [True for i in range(n+1)]
      p = 2
      while (p * p <= n):
  
          # If prime[p] is not
          # changed, then it is a prime
          if (prime[p] == True):
  
              # Update all multiples of p
              for i in range(p * p, n+1, p):
                  prime[i] = False
          p += 1
  
      ans = set()
      for p in range(2, n+1):
          if prime[p]:
              ans.add(p)
  
      return ans
  
  
  MOD = 1000000007
  INT_MAX = inf
  INT_MIN = -inf
  
  
  # Main function
  
  def main():
      t = int(input())
      for _ in range(t):
          n = int(input())
          a = list(map(int, input().split()))
          idxmax = -1
          maxval = -1
          for i in range(n):
              if a[i] >= maxval:
                  maxval = a[i]
                  idxmax = i
  
          if idxmax == n-1:
              print(0)
          else:
              count = 1
              max = a[n-1]
              for i in range(n-2, idxmax, -1):
                  if a[i] > max:
                      max = a[i]
                      count += 1
              print(count)
  
  
  main()
  