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 math
  import sys
  import os
  from io import BytesIO, IOBase
  #from collections import deque, Counter, OrderedDict, defaultdict
  #import heapq
  #ceil,floor,log,sqrt,factorial,pow,pi,gcd
  #import bisect
  #from bisect import bisect_left,bisect_right
  
  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")
  
  def I():
      return(int(input()))
  def lint():
      return(list(map(int,input().split())))
  def insr():
      s = input().strip()
      return(list(s[:len(s)]))
  def invr():
      return(map(int,input().split()))
  import itertools
  from sys import maxsize, stdout, stdin,stderr
  # mod = int(1e9+7)
  import sys
  # def I(): return int(stdin.readline())
  # def lint(): return [int(x) for x in stdin.readline().split()]
  def S(): return list(map(str,input().strip()))
  def grid(r, c): return [lint() for i in range(r)]
  from collections import defaultdict, Counter, deque
  import math
  import heapq
  from heapq import heappop , heappush
  import bisect
  from math import factorial, inf
  from itertools import groupby
  from itertools import permutations as comb
  
  # mod=1000000007
  mod = 998244353
  def gcd(a,b): 
      while b:
          a %= b
          tmp = a
          a = b
          b = tmp
      
      return a
   
  def lcm(a,b): 
      return a  // gcd(a, b) * b
   
  def check_prime(n):
      for i in range(2, int(n ** (1 / 2)) + 1):
          if not n % i:
              return False
      return True
  
  # def nCr(n, r):
   
  #     return (fact(n) // (fact(r)
  #                 * fact(n - r)))
   
  # Returns factorial of n
  # def fact(n):
   
  #     res = 1
       
  #     for i in range(2, n+1):
  #         res = res * i
           
  #     return res
  def primefactors(n):
      num=0
      
      while n % 2 == 0:
          num+=1
          n = n / 2
      
      for i in range(3,int(math.sqrt(n))+1,2):
           
      
          while n % i== 0:
              num+=1
              n = n // i
                 
      
      if n > 2:
          num+=1
      return num
  '''
  def iter_ds(src):
      store=[src]
      while len(store):
          tmp=store.pop()
          if not vis[tmp]:
              vis[tmp]=True
              for j in ar[tmp]:
                  store.append(j)
  '''
  def ask(a,b,c):
      # print('? 1 {}'.format(a),flush=True)
      print(c,a,b,flush=True)
      n=I()
      
      return n
  def linear_sieve(n):
      is_composite=[False]*n
      prime=[]
      for i in range(2,n):
          if not is_composite[i]:
              prime.append(i)
          for j in prime:
              is_composite[i*j]=True
              if i%prime==0:
                  break
      return prime
   
  def dfs(i,p,d):
      
      a,tmp=0,0
      for j in d[i]:
          if j!=p:
              a+=1
              tmp+=dfs(j,i)
      
      if a==0:
          return 0
      
      return tmp/a + 1 
  def primeFactors(n):
      l=[]
      
      while n % 2 == 0:
          l.append(2)
          n = n // 2
      for i in range(3,int(math.sqrt(n))+1,2):
           
          
          while n % i== 0:
              l.append(i)
              n = n // i
  
      if n > 2:
          l.append(n)
      return l
  
  def binpow(a,b,c):
      res=1
      a=a%c
      while b>0:
          if b&1:
              res=(res*a)%c
  
          b=b>>1
          a = (a*a)%c
  
      return res-1
  
  
      
  
  #  Sieve
  d=[]
  primes=[]
  prim=[0]*(10**6+1)
  def sieve(n):
      
      for i in range(n):
          d.append(i)
  
      for i in range(2,n):
  
          if d[i]==i:
              prim[i]=1
              primes.append(i)
          j=0
          while j<len(primes) and primes[j]<=d[i] and primes[j]*i<n:
          
              d[i * primes[j]] = primes[j]
              j+=1
  
          
  def primeFactors(n):
      factors=[]
      while n!=1:
          factors.append(d[n])
          n//=d[n]
      
      return factors
  
  t = 1
  t = I()
  for _ in range(t):
      n=I()
      s = lint()
      curr=s[n-1]
      res=max(s)
      a=0
      for i in range(n-1,-1,-1):
          if curr==res:
              break
          if s[i]>curr:
              curr=s[i]
              a+=1
      print(a)
  
      