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: import os
  import sys
  from io import BytesIO, IOBase
   
  _str = str
  str = lambda x=b"": x if type(x) is bytes else _str(x).encode()
   
  BUFSIZE = 8192
   
   
  def sigma1(a):
      return sum(a)
  def sigma2(a):
      return sum([i*i for i in a])
  def sigma3(a):
      return sum([i*i*i for i in a])
  def prodofpairs(a):
      return (sigma1(a)**2-sigma2(a))//2
  def prodoftriplets(a):
      return (sigma1(a)**3-3*sigma1(a)*sigma2(a)+2*sigma3(a))//6
  from collections import Counter,defaultdict
  from math import gcd
  from collections import deque
  def bfs(graph, start=0):
      used = [False] * len(graph)
      used[start] = True
      q = [start]
      for v in q:
          for w in graph[v]:
              if not used[w]:
                  used[w] = True
                  q.append(w)
   
   
  def layers(graph, start=0):
      used = [False] * len(graph)
      used[start] = True
      q, ret = [start], []
      level=[0]*len(graph)
      while q:
          nq = []
          ret.append(q)
          for v in q:
              for w in graph[v]:
                  if not used[w]:
                      used[w] = True
                      nq.append(w)
                      level[w]=1+level[v]
          q = nq
      return level
   
  import math
   
  # function to find GCD of sub sequence
  # of size k with max GCD in the array
  def findMaxGCD(arr, n, k):
      high = max(arr)
      divisors = [0] * (high + 1)
      for i in range(n) :
          for j in range(1, int(math.sqrt(arr[i])) + 1):
              if (arr[i] % j == 0) :
                  divisors[j] += 1
                  if (j != arr[i] // j):
                      divisors[arr[i] // j] += 1
      for i in range(high, 0, -1):
          if (divisors[i] >= k):
              return i
  def xdist(a,b):
      if(b[0]>=a[0]>=b[2] or b[0]<=a[0]<=b[2] ):
          return 0
      return min(abs(a[0]-b[0]),abs(a[0]-b[2]))
  def ydist(a,b):
      if(b[1]>=a[1]>=b[3] or b[1]<=a[1]<=b[3] ):
          return 0
      return min(abs(a[1]-b[1]),abs(a[1]-b[3]))
  def countdistx(i,j,k,obj):
      m=0
      for i1 in range(k):
          m+=xdist([i,j],obj[i1])
      return m
  def countdisty(i,j,k,obj):
      m=0
      for i1 in range(k):
          m+=ydist([i,j],obj[i1])
      return m
  from bisect import bisect_left
  def dp(i,n,cost,a):
      if(cost==0):
          return 0
      if(cost<0):
          return float("Inf")
      l=[float("Inf")]
      for j in range(i,n+1):
          k=bisect_left(a,a[j]+cost,j)
          if(k<=n and a[k]==a[j]+cost):
              l.append(k-j)
      return min(l)
  def isp(x):
      for i in range(2,int(math.sqrt(x))+1):
          if(x%i==0):
              return False
      return True
  from bisect import bisect_right
  from sys import stdout
  from math import ceil,log2,floor
  def main():
      for _ in range(int(input())):
          n=int(input())
          a=list(map(int,input().split()))
          ans=0
          for i in range(n):
              ans=max(ans,a[i]-i-1)
          print(ans)
  
  
  
   
      
   
   
   
              
   
   
   
      
          
          
              
   
   
   
   
          
   
   
   
   
   
          
   
   
          
   
  main()
   
  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")
   
   
  sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
  input = lambda: sys.stdin.readline().rstrip("\r\n")
   
   
   