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.io.*;
  import java.math.BigInteger;
  
  //code by tishrah_
  public class _practise {
  	
  	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()); 
  		} 
  
  		 int[] ia(int n)
  		{
  			int a[]=new int[n];
  			for(int i=0;i<n;i++)a[i]=nextInt();
  			return a;
  		}
  		 int[][] ia(int n , int m)
  			{
  				int a[][]=new int[n][m];
  				for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextInt();
  				return a;
  			}
  		 long[][] la(int n , int m)
  			{
  				long a[][]=new long[n][m];
  				for(int i=0;i<n;i++) for(int j=0;j<m ;j++) a[i][j]=nextLong();
  				return a;
  			}
  		 char[][] ca(int n , int m)
  			{
  				char a[][]=new char[n][m];
  				for(int i=0;i<n;i++)
  					{
  					String x =next();
  					for(int j=0;j<m ;j++) a[i][j]=x.charAt(j);
  					}
  				return a;
  			}
  
  		long[] la(int n)
  		{
  			long a[]=new long[n];
  			for(int i=0;i<n;i++)a[i]=nextLong();
  			return a;
  		}
  
  		String nextLine() 
  		{ 
  			String str = ""; 
  			try
  			{ 
  				str = br.readLine(); 
  			} 
  			catch (IOException e) 
  			{ 
  				e.printStackTrace(); 
  			} 
  			return str; 
  		} 
  	}
  	static void sort(long[] a) 
  	{int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {long oi=r.nextInt(n), temp=a[i];a[i]=a[(int)oi];a[(int)oi]=temp;}Arrays.sort(a);}
      static void sort(int[] a) 
      {int n=a.length;Random r=new Random();for (int i=0; i<a.length; i++) {int oi=r.nextInt(n), temp=a[i];a[i]=a[oi];a[oi]=temp;}Arrays.sort(a);}
  	public static long sum(long a[])
  	{long sum=0; for(long i : a) sum+=i; return(sum);}
  	public static long count(long a[] , long x)
  	{long c=0; for(long i : a) if(i==x) c++; return(c);}
  	public static int sum(int a[])
  	{ int sum=0; for(int i : a) sum+=i; return(sum);}
  	public static int count(int a[] ,int x)
  	{int c=0; for(int i : a) if(i==x) c++; return(c);}
  	public static int count(String s ,char ch)
  	{int c=0; char x[] = s.toCharArray(); for(char i : x) if(ch==i) c++; return(c);}
  	public static boolean prime(int n)
  	{for(int i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
  	public static boolean prime(long n)
  	{for(long i=2 ; i<=Math.sqrt(n) ; i++) if(n%i==0) return false; return true;}
  	 public static int gcd(int n1, int n2)
  	  { if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
  	 public static long gcd(long n1, long n2)
  	  { if (n2 != 0)return gcd(n2, n1 % n2); else return n1;}
  	 public static int[] freq(int a[], int n)
  	  { int f[]=new int[n+1]; for(int i:a) f[i]++; return f;}
  	 public static int[] pos(int a[], int n)
  	  { int f[]=new int[n+1]; for(int i=0; i<n ;i++) f[a[i]]=i; return f;}
  	 
  	
  	 public static long mindig(long n)
  	 {
  		 long ans=9; while(n>0) { if(n%10<ans) ans=n%10; n/=10; }
  		 return ans;
  	 }
  	 public static long maxdig(long n)
  	 {
  		 long ans=0; while(n>0) { if(n%10>ans) ans=n%10; n/=10; }
  		 return ans;
  	 }
  	  
  	public static boolean palin(String s)
  	{
  		StringBuilder sb = new StringBuilder();
  		sb.append(s);
  		String str=String.valueOf(sb.reverse());
  		if(s.equals(str))
  			return true;
  		else return false;
  	}
  	
  	public static void main(String args[])
  	{
  
  		FastReader in=new FastReader();
  		PrintWriter so = new PrintWriter(new BufferedWriter(new  OutputStreamWriter(System.out)));
  		_practise ob = new _practise();
  		
  	    
  	    int T = in.nextInt();		
  	     //int T = 1;
  		
  	     tc : while(T-->0)
  		 { 	
  	    	int n = in.nextInt();
  	    	int a[] = in.ia(n);
  	    	int ans=0;
  	    	for(int i=0 ; i<n ; i++) if(a[i]>i+1) ans=Math.max(ans,a[i]-(i+1));
  	    	so.println(ans);
  	    	
  		 }
  		
  		
  		so.flush();
  		/*String s = in.next();
  		 * Arrays.stream(f).min().getAsInt()
  		 * BigInteger f = new BigInteger("1"); 1 ke jagah koi bhi value ho skta jo aap 
  		 *                                       initial value banan chahte
  			int a[] = new int[n];
  			 Stack<Integer> stack = new Stack<Integer>();
  			Deque<Integer> q = new LinkedList<>(); or Deque<Integer> q = new ArrayDeque<Integer>();
  			PriorityQueue<Long> pq = new PriorityQueue<Long>();
  			ArrayList<Integer> al = new ArrayList<Integer>(); 
  			StringBuilder sb = new StringBuilder();
  			HashSet<Integer> st = new LinkedHashSet<Integer>();
  			Set<Integer> s = new HashSet<Integer>();
  			Map<Long,Integer> hm = new HashMap<Long, Integer>(); //<key,value>
  			 for(Map.Entry<Integer, Integer> i :hm.entrySet())
  			HashMap<Integer, Integer> hmap = new HashMap<Integer, Integer>();
  			so.println("HELLO");
  			Arrays.sort(a,Comparator.comparingDouble(o->o[0]))
  			 Arrays.sort(a, (aa, bb) -> Integer.compare(aa[1], bb[1]));
  			 Set<String> ts = new TreeSet<>();*/
  	}
  }