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: # Problem: B. Выворачивание массива
  # Contest: Codeforces - Codeforces Round #759 (Div. 2, основан на Отборочном раунде 3 Технокубка 2022)
  # URL: https://codeforces.com/contest/1591/problem/B
  # Memory Limit: 256 MB
  # Time Limit: 1000 ms
  # 
  # Powered by CP Editor (https://cpeditor.org)
  
  import bisect
  
  mod = 10 ** 9 + 7
  eps = 10 ** -9
  
  
  def __gcd(a, b):
  	return a if b == 0 else __gcd(b, a % b)
  	
  
  def __lcm(a, b):
  	return a * b / __gcd(a, b)
  
  
  def __fact(n):
  	return 1 if n == 1 else n * __fact(n - 1)
  
  
  def __mex(a):
  	mex = 0
  	a.sort()
  	for x in a:
  		if x <= mex:
  			mex += 1
  		else:
  			break
  	return mex
  
  def __dist(x1, y1, x2, y2):
  	return (x1 - x2) ** 2 + (y1 - y2) ** 2
  
  def __getprimes(n):
      isprime = [True for i in range(n + 1)]
      primes = []
       
      for i in range(2, n + 1):
          if isprime[i]:
              for j in range(i * i, n + 1, i):
                  isprime[j] = False
      for i in range(2, n + 1):
          if isprime[i]:
              primes.append(i)
      return primes
  
  def __modInverse(a, m):
      
      m0 = m
      y = 0
      x = 1
      if (m == 1):
          return 0
      
      while(a > 1):
      	q = a // m
      	t = m
      	m = a % m
      	a = t
      	t = y
      	y = x - q * y
      	x = t
      if (x < 0):
      	x = x + m0
      return x
      
      
      
  def __isprime(n):
  	if n < 2:
  		return False
  	i = 2
  	while i * i <= n:
  		if n % i == 0:
  			return False
  		i += 1
  	return True
  	
  def __cntprimediv(n):
  	ret = 0
  	x = n
  	i = 2
  	while i * i <= x:
  		while n % i == 0:
  			n //= i
  			ret += 1
  		i += 1
  	if n > 1:
  		ret += 1
  	return ret
  	
  def __primefactors(n):
  	ret = []
  	x = n
  	i = 2
  	while i * i <= x:
  		while n % i == 0:
  			ret.append(i)
  			n //= i
  		i += 1
  	if n > 1:
  		ret.append(n)
  	return ret
  	
  
  def __sumdigit(n):
  	ret = 0
  	while n > 0:
  		ret += n % 10
  		n //= 10
  	return ret
  
  def __zfunc(s):
  	n = len(s)
  	z = [0 for i in range(n)]
  	l = 0
  	r = 0
  	for i in range(1, n):
  		if r >= i:
  			z[i] = min(z[i - l], r - i + 1)
  		while z[i] + i < n and s[z[i]] == s[z[i] + i]:
  			z[i] += 1
  		if i + z[i] - 1 > r:
  			l = i
  			r = i + z[i] - 1
  	return z
  	
  	
  def __to(n, x):
  	ret = ''
  	while n > 0:
  		q = n % x
  		if q < 10:
  			ret += str(q)
  		else:
  			ret += chr(q - 10 + ord('a'))
  		n //= x
  	return ret[::-1]
  	
  	
  def solve(t):
  	n = int(input())
  	a = list(map(int, input().split()))
  	mx = max(a)
  	last = a[-1]
  	ans = 1
  	for i in range(n - 1, -1, -1):
  		if a[i] == mx:
  			break
  		else:
  			if a[i] > last:
  				ans += 1
  				last = a[i]
  	if a[-1] == mx:
  		ans = 0
  	print(ans)
  		
  t = 1
  t = int(input())
  
  
  for i in range(t):
  	solve(i + 1)