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 java.util.*;
  import java.lang.*;
  import java.io.*;
  
  public class A_Era {
  
      public static void main(String[] args) throws java.lang.Exception {
          FastReader sc = new FastReader();
  
          int t = sc.nextInt();
  
          while (t-- > 0) {
  
              int n = sc.nextInt();
              int[] arr = sc.readArray(n);
              long d = 0;
              long md = 0;
              for (int i = 0; i < arr.length; i++) {
                  if ((arr[i] >= 1) && (arr[i] > i + 1)) {
                      d = arr[i] - (i + 1);
  
                      if (d > md) {
                          md = d;
                      }
  
                  }
              }
              System.out.println(md);
          }
  
      }
  
      static class FastReader {
          BufferedReader br;
          StringTokenizer st;
  
          public FastReader() {
              br = new BufferedReader(new InputStreamReader(System.in));
          }
  
          String next() {
              while (st == null || !st.hasMoreElements()) {
                  try {
                      st = new StringTokenizer(br.readLine());
                  } catch (IOException e) {
                      e.printStackTrace();
                  }
              }
              return st.nextToken();
          }
  
          int nextInt() {
              return Integer.parseInt(next());
          }
  
          long nextLong() {
              return Long.parseLong(next());
          }
  
          double nextDouble() {
              return Double.parseDouble(next());
          }
  
          float nextFloat() {
              return Float.parseFloat(next());
          }
  
          String nextLine() {
              String str = "";
              try {
                  str = br.readLine();
              } catch (IOException e) {
                  e.printStackTrace();
              }
              return str;
          }
  
          int[] readArray(int n) {
              int[] a = new int[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextInt();
              return a;
          }
  
          long[] readArrayLong(int n) {
              long[] a = new long[n];
              for (int i = 0; i < n; i++)
                  a[i] = nextLong();
              return a;
          }
  
      }
  
      public static int[] radixSort2(int[] a) {
          int n = a.length;
          int[] c0 = new int[0x101];
          int[] c1 = new int[0x101];
          int[] c2 = new int[0x101];
          int[] c3 = new int[0x101];
          for (int v : a) {
              c0[(v & 0xff) + 1]++;
              c1[(v >>> 8 & 0xff) + 1]++;
              c2[(v >>> 16 & 0xff) + 1]++;
              c3[(v >>> 24 ^ 0x80) + 1]++;
          }
          for (int i = 0; i < 0xff; i++) {
              c0[i + 1] += c0[i];
              c1[i + 1] += c1[i];
              c2[i + 1] += c2[i];
              c3[i + 1] += c3[i];
          }
          int[] t = new int[n];
          for (int v : a)
              t[c0[v & 0xff]++] = v;
          for (int v : t)
              a[c1[v >>> 8 & 0xff]++] = v;
          for (int v : a)
              t[c2[v >>> 16 & 0xff]++] = v;
          for (int v : t)
              a[c3[v >>> 24 ^ 0x80]++] = v;
          return a;
      }
  
      static boolean isPalindrome(String str) {
  
          // Pointers pointing to the beginning
          // and the end of the string
          int i = 0, j = str.length() - 1;
  
          // While there are characters to compare
          while (i < j) {
  
              // If there is a mismatch
              if (str.charAt(i) != str.charAt(j))
                  return false;
  
              // Increment first pointer and
              // decrement the other
              i++;
              j--;
          }
  
          // Given string is a palindrome
          return true;
      }
  
      static boolean palindrome_array(int arr[], int n) {
          // Initialise flag to zero.
          int flag = 0;
  
          // Loop till array size n/2.
          for (int i = 0; i <= n / 2 && n != 0; i++) {
  
              // Check if first and last element are different
              // Then set flag to 1.
              if (arr[i] != arr[n - i - 1]) {
                  flag = 1;
                  break;
              }
          }
  
          // If flag is set then print Not Palindrome
          // else print Palindrome.
          if (flag == 1)
              return false;
          else
              return true;
      }
  
      static boolean allElementsEqual(int[] arr, int n) {
          int z = 0;
          for (int i = 0; i < n - 1; i++) {
              if (arr[i] == arr[i + 1]) {
                  z++;
              }
          }
          if (z == n - 1) {
              return true;
          } else {
              return false;
          }
  
      }
  
      static boolean allElementsDistinct(int[] arr, int n) {
          int z = 0;
          for (int i = 0; i < n - 1; i++) {
              if (arr[i] != arr[i + 1]) {
                  z++;
              }
          }
          if (z == n - 1) {
              return true;
          } else {
              return false;
          }
  
      }
  
      public static void reverse(int[] array) {
  
          // Length of the array
          int n = array.length;
  
          // Swaping the first half elements with last half
          // elements
          for (int i = 0; i < n / 2; i++) {
  
              // Storing the first half elements temporarily
              int temp = array[i];
  
              // Assigning the first half to the last half
              array[i] = array[n - i - 1];
  
              // Assigning the last half to the first half
              array[n - i - 1] = temp;
          }
      }
  
      public static void reverse_Long(long[] array) {
  
          // Length of the array
          int n = array.length;
  
          // Swaping the first half elements with last half
          // elements
          for (int i = 0; i < n / 2; i++) {
  
              // Storing the first half elements temporarily
              long temp = array[i];
  
              // Assigning the first half to the last half
              array[i] = array[n - i - 1];
  
              // Assigning the last half to the first half
              array[n - i - 1] = temp;
          }
      }
  
      static boolean isSorted(int[] a) {
          for (int i = 0; i < a.length - 1; i++) {
              if (a[i] > a[i + 1]) {
                  return false;
              }
          }
  
          return true;
      }
  
      static boolean isReverseSorted(int[] a) {
          for (int i = 0; i < a.length - 1; i++) {
              if (a[i] < a[i + 1]) {
                  return false;
              }
          }
  
          return true;
      }
  
      static int[] rearrangeEvenAndOdd(int arr[], int n) {
          ArrayList<Integer> list = new ArrayList<>();
  
          for (int i = 0; i < n; i++) {
              if (arr[i] % 2 == 0) {
                  list.add(arr[i]);
              }
          }
  
          for (int i = 0; i < n; i++) {
              if (arr[i] % 2 != 0) {
                  list.add(arr[i]);
              }
          }
          int len = list.size();
          int[] array = list.stream().mapToInt(i -> i).toArray();
          return array;
      }
  
      static long[] rearrangeEvenAndOddLong(long arr[], int n) {
          ArrayList<Long> list = new ArrayList<>();
  
          for (int i = 0; i < n; i++) {
              if (arr[i] % 2 == 0) {
                  list.add(arr[i]);
              }
          }
  
          for (int i = 0; i < n; i++) {
              if (arr[i] % 2 != 0) {
                  list.add(arr[i]);
              }
          }
          int len = list.size();
          long[] array = list.stream().mapToLong(i -> i).toArray();
          return array;
      }
  
      static boolean isPrime(long n) {
  
          // Check if number is less than
          // equal to 1
          if (n <= 1)
              return false;
  
          // Check if number is 2
          else if (n == 2)
              return true;
  
          // Check if n is a multiple of 2
          else if (n % 2 == 0)
              return false;
  
          // If not, then just check the odds
          for (long i = 3; i <= Math.sqrt(n); i += 2) {
              if (n % i == 0)
                  return false;
          }
          return true;
      }
  
      static int getSum(int n) {
          int sum = 0;
  
          while (n != 0) {
              sum = sum + n % 10;
              n = n / 10;
          }
  
          return sum;
      }
  
      static int gcd(int a, int b) {
          if (b == 0)
              return a;
          return gcd(b, a % b);
      }
  
      static long gcdLong(long a, long b) {
          if (b == 0)
              return a;
          return gcdLong(b, a % b);
      }
  
      static void swap(int[] arr, int i, int j) {
          int temp = arr[i];
          arr[i] = arr[j];
          arr[j] = temp;
      }
  
      static int countDigit(long n) {
          return (int) Math.floor(Math.log10(n) + 1);
      }
  
      static int sumFromString(String s) {
          int sum = 0;
          for (int i = 0; i < s.length(); i++) {
              if (Character.isDigit(s.charAt(i))) {
                  sum = sum + Character.getNumericValue(s.charAt(i));
              }
          }
          return sum;
      }
  
      static int[] freqCounter(int arr[]) {
          int[] fr = new int[arr.length];
          int[] fi = new int[arr.length];
          int visited = -1;
          for (int i = 0; i < arr.length; i++) {
              int count = 1;
              for (int j = i + 1; j < arr.length; j++) {
                  if (arr[i] == arr[j]) {
                      count++;
                      // To avoid counting same element again
                      fr[j] = visited;
                  }
              }
              if (fr[i] != visited)
                  fr[i] = count;
          }
          for (int i = 0; i < fr.length; i++) {
              if (fr[i] != visited)
                  fi[i] = fr[i];
  
          }
          return fi;
      }
  
  }