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 copy
  import math
  
  fin = None
  fout = None
  uabet = ['A', 'B', 'C', 'D', 'E', 'F', 'G', 'H', 'I', 'J', 'K', 'L', 'M', 'N', 'O', 'P', 'Q', 'R', 'S', 'T', 'U', 'V',
           'W', 'X', 'Y', 'Z']
  abet = ['a', 'b', 'c', 'd', 'e', 'f', 'g', 'h', 'i', 'j', 'k', 'l', 'm', 'n', 'o', 'p', 'q', 'r', 's', 't', 'u', 'v',
          'w', 'x', 'y', 'z']
  INF = 10 ** 18
  
  
  def start(filename):
      filename = filename.replace('.py', '')
      global fin
      global fout
      infile = filename + ".in"
      outfile = filename + ".out"
      fin = open(infile)
      fout = open(outfile, "w")
  
  
  def permutation(lst):
      if len(lst) == 0:
          return []
      if len(lst) == 1:
          return [lst]
      l = []
      for i in range(len(lst)):
          m = lst[i]
          remLst = lst[:i] + lst[i + 1:]
          for p in permutation(remLst):
              l.append([m] + p)
      return l
  
  
  def subsets(x):
      retset = []
      if len(x) == 0:
          retset = [[]]
      else:
          first = x[-1]
          cop = x.copy()
          cop.remove(first)
          y = subsets(cop)
          retset.extend(copy.deepcopy(y))
          m = copy.deepcopy(y)
          for i in range(len(y)):
              k = m[0]
              m.remove(k)
              k.append(first)
              m.append(k)
          retset.extend(m.copy())
      return retset
  
  
  class Point:
      def __init__(self, x, y):
          self.x = x
          self.y = y
  
      def __str__(self):
          return str(self.x) + str(self.y)
  
      def __hash__(self):
          return hash(str(self))
  
      def __eq__(self, other):
          return self.x == other.x and self.y == other.y
  
  
  class frac:
      def __init__(self, n, d):
          self.n = n
          self.d = d
  
      def __str__(self):
          return str(self.n) + '/' + str(self.d)
  
      def __hash__(self):
          return hash(str(self))
  
      def simp(self):
          div = math.gcd(self.n, self.d)
          self.n = self.n / div
          self.d = self.d / div
  
      def __eq__(self, other):
          return self.n / self.d == other.n / other.d
  
      def __lt__(self, other):
          return self.n / self.d < other.n / other.d
  
  
  class Interval:
      def __init__(self, start, end):
          self.start = start
          self.end = end
  
      def __str__(self):
          return str(self.start) + str(self.end)
  
      def __hash__(self):
          return hash(str(self))
  
      def __eq__(self, other):
          return self.start == other.start and self.end == other.end
  
      def __lt__(self, other):
          return self.start < other.start
  
  
  def overlap(verl1, verl2):
      if verl1.end >= verl2.start:
          return True
      else:
          return False
  
  
  def merge(verl1, verl2):
      retverl = Interval(min(verl1.start, verl2.start), max(verl1.end, verl2.end))
      return retverl
  
  
  def changeBase(num, base):
      if base == 2:
          return "{0:b}".format(num)
      power = math.floor(math.log(num, base))
      remainder = num
      end = []
      iter = list(range(power + 1))
      iter.reverse()
      for i in iter:
          if remainder >= base ** i:
              q, r = divmod(remainder, base ** i)
              end.append(q)
              remainder = r
          else:
              end.append(0)
      retlist = []
      for i in end:
          if i > 9:
              retlist.append(uabet[i - 10])
          else:
              retlist.append(i)
      retlist = map(str, retlist)
      return ''.join(retlist)
  
  
  def prime(num):
      if num == 2:
          return True
      if num % 2 == 0:
          return False
      for i in range(2, math.ceil(num ** 0.5) + 1):
          if num % i == 0:
              return False
      return True
  
  
  def solve(n,a):
      t = 0
      for i in range(n):
          if a[i]-(i+1+t)>0:
              t+=a[i]-(i+1+t)
      return t
  
  def main():
      t = int(input())
      for i in range(t):
          n = int(input())
          a = list(map(int,input().split(' ')))
          print(solve(n,a))
  
  
  main()
  