Problem: You're given an array a_1, …, a_n of n non-negative integers.

Let's call it sharpened if and only if there exists an integer 1 ≤ k ≤ n such that a_1 < a_2 < … < a_k and a_k > a_{k+1} > … > a_n. In particular, any strictly increasing or strictly decreasing array is sharpened. For example:

  * The arrays [4], [0, 1], [12, 10, 8] and [3, 11, 15, 9, 7, 4] are sharpened; 
  * The arrays [2, 8, 2, 8, 6, 5], [0, 1, 1, 0] and [2, 5, 6, 9, 8, 8] are not sharpened. 



You can do the following operation as many times as you want: choose any strictly positive element of the array, and decrease it by one. Formally, you can choose any i (1 ≤ i ≤ n) such that a_i>0 and assign a_i := a_i - 1.

Tell if it's possible to make the given array sharpened using some number (possibly zero) of these operations.

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 15\ 000) — the number of test cases. The description of the test cases follows.

The first line of each test case contains a single integer n (1 ≤ n ≤ 3 ⋅ 10^5).

The second line of each test case contains a sequence of n non-negative integers a_1, …, a_n (0 ≤ a_i ≤ 10^9).

It is guaranteed that the sum of n over all test cases does not exceed 3 ⋅ 10^5.

Output

For each test case, output a single line containing "Yes" (without quotes) if it's possible to make the given array sharpened using the described operations, or "No" (without quotes) otherwise.

Example

Input


10
1
248618
3
12 10 8
6
100 11 15 9 7 8
4
0 1 1 0
2
0 0
2
0 1
2
1 0
2
1 1
3
0 1 0
3
1 0 1


Output


Yes
Yes
Yes
No
No
Yes
Yes
Yes
Yes
No

Note

In the first and the second test case of the first test, the given array is already sharpened.

In the third test case of the first test, we can transform the array into [3, 11, 15, 9, 7, 4] (decrease the first element 97 times and decrease the last element 4 times). It is sharpened because 3 < 11 < 15 and 15 > 9 > 7 > 4.

In the fourth test case of the first test, it's impossible to make the given array sharpened.

Solution (in PYTHON):

Solution:
import sys 
# sys.setrecursionlimit(10**6) 
from sys import stdin, stdout
import bisect            #c++ upperbound
import math
import heapq
def modinv(n,p):
    return pow(n,p-2,p)
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())
import math  
def Divisors(n) : 
    l = []  
    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)
                l.append(n//i)
    return l

"""*******************************************************"""
def main():
    t=inin()
    for _ in range(t):
        n=inin()
        a=ain()
        j=0
        for i in range(n):
            if(a[i]>=i):
                j=i
            else:
                break
        a=a[::-1]
        k=0
        for i in range(n):
            if(a[i]>=i):
                k=i 
            else:
                break
        # print(j,k)
        if(j+1+k+1<=n):
            print("No")
        else:
            print("Yes")
######## Python 2 and 3 footer by Pajenegod and c1729
 
# Note because cf runs old PyPy3 version which doesn't have the sped up
# unicode strings, PyPy3 strings will many times be slower than pypy2.
# There is a way to get around this by using binary strings in PyPy3
# but its syntax is different which makes it kind of a mess to use.
 
# So on cf, use PyPy2 for best string performance.
 
py2 = round(0.5)
if py2:
    from future_builtins import ascii, filter, hex, map, oct, zip
    range = xrange
 
import os, sys
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
import sys
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'0' [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'0' [0]:
        A.append(sign*numb)
    return A
 
if __name__== "__main__":
  main()