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>
# ===============================================================================================
# importing some useful libraries.
from __future__ import division, print_function
from fractions import Fraction
import sys
import os
from io import BytesIO, IOBase
from itertools import *
import bisect
from heapq import *
from math import ceil, floor
from copy import *
from collections import deque, defaultdict
from collections import Counter as counter  # Counter(list)  return a dict with {key: count}
from itertools import combinations  # if a = [1,2,3] then print(list(comb(a,2))) -----> [(1, 2), (1, 3), (2, 3)]
from itertools import permutations as permutate
from bisect import bisect_left as bl
from operator import *
# If the element is already present in the list,

# the left most position where element has to be inserted is returned.
from bisect import bisect_right as br
from bisect import bisect

# If the element is already present in the list,
# the right most position where element has to be inserted is returned

# ==============================================================================================
# fast I/O region

BUFSIZE = 8192
from sys import stderr


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)

# inp = lambda: sys.stdin.readline().rstrip("\r\n")

# ===============================================================================================
### START ITERATE RECURSION ###
from types import GeneratorType


def iterative(f, stack=[]):
    def wrapped_func(*args, **kwargs):
        if stack: return f(*args, **kwargs)
        to = f(*args, **kwargs)
        while True:
            if type(to) is GeneratorType:
                stack.append(to)
                to = next(to)
                continue
            stack.pop()
            if not stack: break
            to = stack[-1].send(to)
        return to

    return wrapped_func


#### END ITERATE RECURSION ####

# ===============================================================================================
# some shortcuts

mod = 1000000007


def inp(): return sys.stdin.readline().rstrip("\r\n")  # for fast input


def out(var): sys.stdout.write(str(var))  # for fast output, always take string


def lis(): return list(map(int, inp().split()))


def stringlis(): return list(map(str, inp().split()))


def sep(): return map(int, inp().split())


def strsep(): return map(str, inp().split())


def fsep(): return map(float, inp().split())


def nextline(): out("\n")  # as stdout.write always print sring.


def testcase(t):
    for p in range(t):
        solve()


def pow(x, y, p):
    res = 1  # Initialize result
    x = x % p  # Update x if it is more , than or equal to p
    if (x == 0):
        return 0
    while (y > 0):
        if ((y & 1) == 1):  # If y is odd, multiply, x with result
            res = (res * x) % p

        y = y >> 1  # y = y/2
        x = (x * x) % p
    return res


from functools import reduce


def factors(n):
    return set(reduce(list.__add__,
                      ([i, n // i] for i in range(1, int(n ** 0.5) + 1) if n % i == 0)))


def gcd(a, b):
    if a == b: return a
    while b > 0: a, b = b, a % b
    return a


# discrete binary search
# minimise:
# def search():
#     l = 0
#     r = 10 ** 15
#
#     for i in range(200):
#         if isvalid(l):
#             return l
#         if l == r:
#             return l
#         m = (l + r) // 2
#         if isvalid(m) and not isvalid(m - 1):
#             return m
#         if isvalid(m):
#             r = m + 1
#         else:
#             l = m
#     return m

# maximise:
# def search():
#     l = 0
#     r = 10 ** 15
#
#     for i in range(200):
#         # print(l,r)
#         if isvalid(r):
#             return r
#         if l == r:
#             return l
#         m = (l + r) // 2
#         if isvalid(m) and not isvalid(m + 1):
#             return m
#         if isvalid(m):
#             l = m
#         else:
#             r = m - 1
#     return m


#to find factorial and ncr
# N=100000
# mod = 10**9 +7
# fac = [1, 1]
# finv = [1, 1]
# inv = [0, 1]
#
# for i in range(2, N + 1):
#     fac.append((fac[-1] * i) % mod)
#     inv.append(mod - (inv[mod % i] * (mod // i) % mod))
#     finv.append(finv[-1] * inv[-1] % mod)
#
#
# def comb(n, r):
#     if n < r:
#         return 0
#     else:
#         return fac[n] * (finv[r] * finv[n - r] % mod) % mod


##############Find sum of product of subsets of size k in a array
# ar=[0,1,2,3]
# k=3
# n=len(ar)-1
# dp=[0]*(n+1)
# dp[0]=1
# for pos in range(1,n+1):
#     dp[pos]=0
#     l=max(1,k+pos-n-1)
#     for j in range(min(pos,k),l-1,-1):
#         dp[j]=dp[j]+ar[pos]*dp[j-1]
# print(dp[k])

def prefix_sum(ar):  # [1,2,3,4]->[1,3,6,10]
    return list(accumulate(ar))


def suffix_sum(ar):  # [1,2,3,4]->[10,9,7,4]
    return list(accumulate(ar[::-1]))[::-1]


def N():
    return int(inp())


# =========================================================================================
from collections import defaultdict



def numberOfSetBits(i):
    i = i - ((i >> 1) & 0x55555555)
    i = (i & 0x33333333) + ((i >> 2) & 0x33333333)
    return (((i + (i >> 4) & 0xF0F0F0F) * 0x1010101) & 0xffffffff) >> 24



def solve():
    n=N()
    ar=lis()
    cl=deque([ar[0]])
    j=1
    h=0
    nl=[]
    while(cl):
        v=cl.popleft()
        # print(v,j)
        # print(cl)
        # if 1:
        #     print(nl)


        if j<n:
            m=ar[j]
            nl.append(m)
            j+=1
        while(j<n and ar[j]>m):
            m=ar[j]
            nl.append(ar[j])
            j+=1
        if not cl:
            cl=deque(nl)
            if nl:
                # print(nl)
                nl=[]
                h+=1
    if nl:
        h+=1

    print(h)



# 1
# 11
# 1 2 3 4 6 7 5 8 9 10 11







    

    


#solve()
testcase(int(inp()))

