human
Can you solve the below in PYTHON3?
Shohag has an integer sequence a_1, a_2, …, a_n. He can perform the following operation any number of times (possibly, zero):

  * Select any positive integer k (it can be different in different operations). 
  * Choose any position in the sequence (possibly the beginning or end of the sequence, or in between any two elements) and insert k into the sequence at this position. 
  * This way, the sequence a changes, and the next operation is performed on this changed sequence. 


For example, if a=[3,3,4] and he selects k = 2, then after the operation he can obtain one of the sequences [\underline{2},3,3,4], [3,\underline{2},3,4], [3,3,\underline{2},4], or [3,3,4,\underline{2}].

Shohag wants this sequence to satisfy the following condition: for each 1 ≤ i ≤ |a|, a_i ≤ i. Here, |a| denotes the size of a.

Help him to find the minimum number of operations that he has to perform to achieve this goal. We can show that under the constraints of the problem it's always possible to achieve this goal in a finite number of operations.

Input

The first line contains a single integer t (1 ≤ t ≤ 200) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 100) — the initial length of the sequence.

The second line of each test case contains n integers a_1, a_2, …, a_n (1 ≤ a_i ≤ 10^9) — the elements of the sequence.

Output

For each test case, print a single integer — the minimum number of operations needed to perform to achieve the goal mentioned in the statement.

Example

Input

4
3
1 3 4
5
1 2 5 7 4
1
1
3
69 6969 696969

Output

1
3
0
696966

Note

In the first test case, we have to perform at least one operation, as a_2=3>2. We can perform the operation [1, 3, 4] → [1, \underline{2}, 3, 4] (the newly inserted element is underlined), now the condition is satisfied.

In the second test case, Shohag can perform the following operations:

[1, 2, 5, 7, 4] → [1, 2, \underline{3}, 5, 7, 4] → [1, 2, 3, \underline{4}, 5, 7, 4] → [1, 2, 3, 4, 5, \underline{3}, 7, 4].

In the third...
 Give me a Python program to find the minimum number of operations that he has to perform to achieve this goal?
assistant
#Here is Python program to find the minimum number of operations that he has to perform to achieve this goal.

#!/usr/bin/env python

from __future__ import division, print_function
import os
import sys
from fileinput import input
from io import BytesIO, IOBase

import math
# import random
# import bisect
# import collections
# import heapq
# from collections import deque
# from functools import reduce

inf=float("inf")
# mod=1000000007

if sys.version_info[0] < 3:
    from __builtin__ import xrange as range
    from future_builtins import ascii, filter, hex, map, oct, zip

#   n = int(input())
#   arr=list (map(int, input().split()))

def main():
    def solve():
        n=int(input())
        arr =list(map(int, input().split()))
        ans=0
        for i in range(n):
            if arr[i]>i+1+ans:
                ans+=(arr[i]-1-i-ans)
                # print(i,ans)
        return (ans)

    t = int(input())
    for i in range(t):
        print(solve())
        # solve()

# region fastio

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),...