description: 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 test case, the sequence already satisfies the condition.
solution: from bisect import insort,bisect_right,bisect_left
  from sys import stdout, stdin, setrecursionlimit
  from heapq import heappush, heappop, heapify 
  from io import BytesIO, IOBase
  from collections import *
  from itertools import *
  from random import * 
  from string import *
  from queue import *
  from math import *
  from re import *
  from os import *
  # sqrt,ceil,floor,factorial,gcd,log2,log10,comb
  from operator import truediv
  
  ####################################---fast-input-output----#########################################
  
  
  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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
              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 = read(self._fd, max(fstat(self._fd).st_size, 8192))
              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:
              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")
  
  
  stdin, stdout = IOWrapper(stdin), IOWrapper(stdout)
  graph, mod, szzz =  {}, 10**9 + 7, lambda: sorted(zzz())
  def getStr(): return input()
  def getInt(): return int(input())
  def listStr(): return list(input())
  def getStrs(): return input().split()
  def isInt(s): return '0' <= s[0] <= '9'
  def input(): return stdin.readline().strip()
  def zzz(): return [int(i) for i in input().split()]
  def output(answer, end='\n'): stdout.write(str(answer) + end)
  def lcd(xnum1, xnum2): return (xnum1 * xnum2 // gcd(xnum1, xnum2))
  
  
  def getPrimes(N = 10**5):
      SN = int(sqrt(N))
      sieve = [i for i in range(N+1)]
      sieve[1] = 0
      for i in sieve:
          if i > SN:
              break
          if i == 0:
              continue
          for j in range(2*i, N+1, i):
              sieve[j] = 0
      prime = [i for i in range(N+1) if sieve[i] != 0] 
      return prime
  def primeFactor(n,prime=getPrimes()):
      lst = []
      mx=int(sqrt(n))+1
      for i in prime:
          if i>mx:break
          while n%i==0:
              lst.append(i)
              n//=i
      if n>1:
          lst.append(n)
      return lst    
  
  dx = [-1, 1, 0, 0, 1, -1, 1, -1]
  dy = [0, 0, 1, -1, 1, -1, -1, 1]
  daysInMounth = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
  
  
  #################################################---Some Rule For Me To Follow---#################################
  """
      --instants of Reading problem continuously try to understand them.
  
      --Try & again try, maybe you're just one statement away!
  
  """
  ##################################################---START-CODING---###############################################
  
  
  num = getInt()
  
  
  for _ in range(num):
      n = getInt()
  
      arr = [0]+zzz()
  
      ans = 0
  
  
      for i in range(1,n+1):
          x = arr[i]
          if i+ans<x:
              ans+= (x-(i+ans))
  
      print(ans)
  
  