Monocarp had a tree which consisted of n vertices and was rooted at vertex 1. He decided to study BFS ([Breadth-first search](https://en.wikipedia.org/wiki/Breadth-first_search)), so he ran BFS on his tree, starting from the root. BFS can be described by the following pseudocode:
    
    
    a = [] # the order in which vertices were processed  
    q = Queue()  
    q.put(1) # place the root at the end of the queue  
    while not q.empty():  
        k = q.pop() # retrieve the first vertex from the queue  
        a.append(k) # append k to the end of the sequence in which vertices were visited  
        for y in g[k]: # g[k] is the list of all children of vertex k, sorted in ascending order  
            q.put(y)  
    

Monocarp was fascinated by BFS so much that, in the end, he lost his tree. Fortunately, he still has a sequence of vertices, in which order vertices were visited by the BFS algorithm (the array a from the pseudocode). Monocarp knows that each vertex was visited exactly once (since they were put and taken from the queue exactly once). Also, he knows that all children of each vertex were viewed in ascending order.

Monocarp knows that there are many trees (in the general case) with the same visiting order a, so he doesn't hope to restore his tree. Monocarp is okay with any tree that has minimum height.

The height of a tree is the maximum depth of the tree's vertices, and the depth of a vertex is the number of edges in the path from the root to it. For example, the depth of vertex 1 is 0, since it's the root, and the depth of all root's children are 1.

Help Monocarp to find any tree with given visiting order a and minimum height.

Input

The first line contains a single integer t (1 ≤ t ≤ 1000) — the number of test cases.

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of vertices in the tree.

The second line of each test case contains n integers a_1, a_2, ..., a_n (1 ≤ a_i ≤ n; a_i ≠ a_j; a_1 = 1) — the order in which the vertices were visited by the BFS algorithm.

It's guaranteed that the total sum of n over test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case print the minimum possible height of a tree with the given visiting order a.

Example

Input


3
4
1 4 3 2
2
1 2
3
1 2 3


Output


3
1
1

Note

In the first test case, there is only one tree with the given visiting order: 

<image>

In the second test case, there is only one tree with the given visiting order as well: 

<image>

In the third test case, an optimal tree with the given visiting order is shown below: 

<image>
from __future__ import division, print_function
import threading,bisect,math,heapq,sys
# threading.stack_size(2**27)
# sys.setrecursionlimit(10**4)
from sys import stdin, stdout
i_m=9223372036854775807    
def cin():
    return map(int,sin().split())
def ain():                           #takes array as input
    return list(map(int,sin().split()))
def sin():
    return input()
def inin():
    return int(input()) 
prime=[]
def dfs(n,d,v,a,c):
    v[n]=1
    if n in d:
        x=d[n]
    else:
        a[n]+=c
        x=[]
    p=a[n]
    for i in x:
        if i not in v:
            dfs(i,d,v,a,c)
            p+=a[i]
    a[n]=p
    return p

"""**************************MAIN*****************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=ain()
        i=0
        j=i+1
        v=[-1]*n
        v[0]=0
        while(i<n):
            while(j<n-1 and a[j]<a[j+1]):
                v[j]=v[i]+1
                j+=1
            v[j]=v[i]+1
            j+=1
            i+=1
            if(j==n):
                break
        print(v[-1])
        
"""*******************************************************"""

######## Python 2 and 3 footer by Pajenegod and c1729
fac=[]
def fact(n,mod):
    global fac
    fac.append(1)
    for i in range(1,n+1):
        fac.append((fac[i-1]*i)%mod)
    f=fac[:]
    return f
def nCr(n,r,mod):
    global fac
    x=fac[n]
    y=fac[n-r]
    z=fac[r]
    x=moddiv(x,y,mod)
    return moddiv(x,z,mod)
def moddiv(m,n,p):
    x=pow(n,p-2,p)
    return (m*x)%p
def GCD(x, y): 
    x=abs(x)
    y=abs(y)
    if(min(x,y)==0):
        return max(x,y)
    while(y): 
        x, y = y, x % y 
    return x 
def Divisors(n) : 
    l = []  
    ll=[]
    for i in range(1, int(math.sqrt(n) + 1)) :
        if (n % i == 0) : 
            if (n // i == i) : 
                l.append(i) 
            else : 
                l.append(i)
                ll.append(n//i)
    l.extend(ll[::-1])
    return l
def SieveOfEratosthenes(n): 
    global prime
    prime = [True for i in range(n+1)] 
    p = 2
    while (p * p <= n): 
        if (prime[p] == True): 
            for i in range(p * p, n+1, p): 
                prime[i] = False
        p += 1
    f=[]
    for p in range(2, n): 
        if prime[p]: 
            f.append(p)
    return f
def primeFactors(n): 
    a=[]
    while n % 2 == 0: 
        a.append(2) 
        n = n // 2 
    for i in range(3,int(math.sqrt(n))+1,2):  
        while n % i== 0: 
            a.append(i) 
            n = n // i  
    if n > 2: 
        a.append(n)
    return a
"""*******************************************************"""
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
import os
from io import IOBase, BytesIO
BUFSIZE = 8192
class FastIO(BytesIO):
    newlines = 0
 
    def __init__(self, file):
        self._file = file
        self._fd = file.fileno()
        self.writable = "x" in file.mode or "w" in file.mode
        self.write = super(FastIO, self).write if self.writable else None
 
    def _fill(self):
        s = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
        self.seek((self.tell(), self.seek(0,2), super(FastIO, self).write(s))[0])
        return s
 
    def read(self):
        while self._fill(): pass
        return super(FastIO,self).read()
 
    def readline(self):
        while self.newlines == 0:
            s = self._fill(); self.newlines = s.count(b"\n") + (not s)
        self.newlines -= 1
        return super(FastIO, self).readline()
 
    def flush(self):
        if self.writable:
            os.write(self._fd, self.getvalue())
            self.truncate(0), self.seek(0)
 
class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        if py2:
            self.write = self.buffer.write
            self.read = self.buffer.read
            self.readline = self.buffer.readline
        else:
            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')
 
sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip('\r\n') 
# Cout implemented in Python
class ostream:
    def __lshift__(self,a):
        sys.stdout.write(str(a))
        return self
cout = ostream()
endl = '\n'
 
# Read all remaining integers in stdin, type is given by optional argument, this is fast
def readnumbers(zero = 0):
    conv = ord if py2 else lambda x:x
    A = []; numb = zero; sign = 1; i = 0; s = sys.stdin.buffer.read()
    try:
        while True:
            if s[i] >= b'R' [0]:
                numb = 10 * numb + conv(s[i]) - 48
            elif s[i] == b'-' [0]: sign = -1
            elif s[i] != b'\r' [0]:
                A.append(sign*numb)
                numb = zero; sign = 1
            i += 1
    except:pass
    if s and s[-1] >= b'R' [0]:
        A.append(sign*numb)
    return A
 
# threading.Thread(target=main).start()
if __name__== "__main__":
  main()