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.
1
39
481199252 336470888 634074578 642802746 740396295 773386884 579721198 396628655 503722503 971207868 202647942 2087506 268792718 46761498 443917727 16843338 125908043 691952768 717268783 787375312 150414369 693319712 519096230 45277106 856168102 762263554 674936674 407246545 274667941 279198849 527268921 155191459 421436316 536134606 286802932 140515506 646837311 233465964 451394766

971207858

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import static java.lang.Math.*;
import static java.lang.System.out;
import java.util.*;
import java.io.File;
import java.io.PrintStream;
import java.io.PrintWriter;
import java.math.BigInteger;
public class Main {
 
	
	/* 10^(7) = 1s.
	 * ceilVal = (a+b-1) / b */
	
	static final int mod = 1000000007;
	static final long temp = 998244353;
	static final long MOD = 1000000007;
	static final long M = (long)1e9+7;
 
	static class Pair implements Comparable<Pair> 
	{
		int first, second;
		
		public Pair(int first, int second) 
		{
			this.first = first;
			this.second = second;
		}
		
		public int compareTo(Pair ob) 
		{
			return (int)(first - ob.first);
		}
	}
	
	
	static class Tuple implements Comparable<Tuple>
	{
		int first, second,third;
		public Tuple(int first, int second, int third)
		{
			this.first = first;
			this.second = second;
			this.third = third;
		}
		public int compareTo(Tuple o)
		{
			return (int)(o.third - this.third);
		}
	}
	
	public static class DSU
	{
		int[] parent;
		int[] rank; //Size of the trees is used as the rank
		public DSU(int n)
		{
			parent = new int[n];
			rank = new int[n];
			Arrays.fill(parent, -1);
			Arrays.fill(rank, 1);
		}
		
		public int find(int i) //finding through path compression
		{
			return parent[i] < 0 ? i : (parent[i] = find(parent[i]));
		}
		
		public boolean union(int a, int b) //Union Find by Rank 
		{
			a = find(a);
			b = find(b);
			
			if(a == b) return false; //if they are already connected we exit by returning false.
			
			// if a's parent is less than b's parent
			if(rank[a] < rank[b])
			{
				//then move a under b
				parent[a] = b;
			}
			//else if rank of j's parent is less than i's parent
			else if(rank[a] > rank[b])
			{
				//then move b under a
				parent[b] = a;
			}
			//if both have the same rank.
			else
			{
				//move a under b (it doesnt matter if its the other way around.
				parent[b] = a;
				rank[a] = 1 + rank[a];
			}
			return true;
		}	
	}

	static class Reader 
	{
		BufferedReader br=new BufferedReader(new InputStreamReader(System.in));
		StringTokenizer st=new StringTokenizer("");
		String next() {
			while (!st.hasMoreTokens())
				try {
					st=new StringTokenizer(br.readLine());
				} catch (IOException e) {
					e.printStackTrace();
				}
			return st.nextToken();
		}
		
		int nextInt() {
			return Integer.parseInt(next());
		}
		
		int[] readArray(int n) throws IOException {
			int[] a=new int[n];
			for (int i=0; i<n; i++) a[i]=nextInt();
			return a;
		}
		
		long[] longReadArray(int n) throws IOException {
			long[] a=new long[n];
			for (int i=0; i<n; i++) a[i]=nextLong();
			return a;
		}
		
		long nextLong() {
			return Long.parseLong(next());
		}
		double nextDouble() {
			return Double.parseDouble(next());
		}
	}
	
	public static int gcd(int a, int b)
	{
		if(b == 0)
		 return a;
		
		else
		return gcd(b,a%b);
	} 
 
		public static long lcm(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
	public static long LongGCD(long a, long b)
	{
		if(b == 0)
			 return a;
			
			else
			return LongGCD(b,a%b);
	}
	
		public static long LongLCM(long a, long b)
	    {
	        return (a / LongGCD(a, b)) * b;
	    }
	
		//Count the number of coprime's upto N
		public static long phi(long n)  //euler totient/phi function
	    { 
	       long ans = n;
//	       for(long i = 2;i*i<=n;i++)
//	       {
//	    	   if(n%i == 0)
//	    	   {
//	    		   while(n%i == 0) n/=i;
//	    		   ans -= (ans/i);
//	    	   }
//	       }
//	       
//	       if(n > 1)
//	       {
//	    	   ans -= (ans/n);
//	       }
	       
	       for(long i = 2;i<=n;i++)
	       {
	    	   if(isPrime(i))
	    	   {
	    		   ans -= (ans/i);
	    	   }
	       }
	       
	       return ans;
	       
	    }	
	
	public static long fastPow(long x, long n) 
	{
		if(n == 0)
			return 1;
		else if(n%2 == 0)
			return fastPow(x*x,n/2);
		else
			return x*fastPow(x*x,(n-1)/2);
	}
	
	   public static long modPow(long x, long y, long p)
	    {
	        long res = 1;
	        x = x % p;
	 
	        while (y > 0) {
	            if (y % 2 == 1)
	                res = (res * x) % p;
	 
	            y = y >> 1; 
	            x = (x * x) % p;
	        }
	        return res;
	    }
	   
	   static long modInverse(long n, long p)
	    {
	        return modPow(n, p - 2, p);
	    }
	 
	    // Returns nCr % p using Fermat's little theorem.
	    
	    public static long nCrModP(long n, long r,long p)
	    {
	    	if (n<r) 
	              return 0;
	          
	        if (r == 0)
	            return 1;
	 
	        long[] fac = new long[(int)(n) + 1];
	        fac[0] = 1;
	 
	        for (int i = 1; i <= n; i++)
	            fac[i] = fac[i - 1] * i % p;
	 
	        return (fac[(int)(n)] * modInverse(fac[(int)(r)], p)
	                % p * modInverse(fac[(int)(n - r)], p)
	                % p)
	            % p;
	    }
	    
	    public static long fact(long n) 
	    {
	    	  long[] fac = new long[(int)(n) + 1];
		        fac[0] = 1;
		 
		        for (long i = 1; i <= n; i++)
		            fac[(int)(i)] = fac[(int)(i - 1)] * i;
		        
		        return fac[(int)(n)];
		}
	    
	    public static long nCr(long n, long k) 
	    {
			long ans = 1;
			for(long i = 0;i<k;i++)
			{
				ans *= (n-i);
				ans /= (i+1);
			}
			return ans;
		}
 
		//Modular Operations for Addition and Multiplication. 
		   public static long perfomMod(long x)
		   {
		        return ((x%M + M)%M);
		    }
		   public static long addMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)+perfomMod(b));
		    }
		  
		   public static long subMod(long a, long b)
		   {
		        return perfomMod(perfomMod(a)-perfomMod(b));
		   }
		   
		   public static long mulMod(long  a, long b)
		   {
		        return perfomMod(perfomMod(a)*perfomMod(b));
		   }
		   
		   public static boolean isPrime(long n) 
			{
				if(n == 1)
				{
					return false;
				}
				
				//check only for sqrt of the number as the divisors 
				//keep repeating so only half of them are required. So,sqrt.
				for(int i = 2;i*i<=n;i++)
				{
					if(n%i == 0)
					{
						return false;
					}
				}
				return true;
			}
			
			public static List<Long> SieveList(int n)
			{
				boolean prime[] = new boolean[(int)(n+1)];
				Arrays.fill(prime, true);
				
				List<Long> l = new ArrayList<>();
				for (long p = 2; p*p<=n; p++) 
				{ 
					if (prime[(int)(p)] == true) 
					{ 	
					    for(long i = p*p; i<=n; i += p) 
						{
						    prime[(int)(i)] = false; 
						}	
					} 
				} 
		 
				for (long p = 2; p<=n; p++) 
				{
				    if (prime[(int)(p)] == true)
				    {
				       l.add(p); 
				    }
				}
				
				return l;
			}

			public static long log2(long n)
		    {
				 long ans = (long)(log(n)/log(2));
				 return ans;
			}
			   
			public static boolean isPow2(long n)
			{
				 return (n  != 0 && ((n & (n-1))) == 0);
			}	
		   
		   public static boolean isSq(int x)  
		   {
			    long s = (long)Math.round(Math.sqrt(x));
			    return s*s==x;
		   }
		
			   /*
			    * 
			    * >=         <=
			   0 1 2 3 4 5 6 7  
			   5 5 5 6 6 6 7 7
			   lower_bound for 6 at index 3 (>=)
			   upper_bound for 6 at index 6(To get six reduce by one) (<=)
			   */

		   public static int LowerBound(long a[], long x) 
		   {
			      int l=-1,r=a.length;
				  while(l+1<r) 
				  {
				    int m=(l+r)>>>1;
				    if(a[m]>=x) r=m;
				    else l=m;
				  }
				  return r;
			}
		   
			public static int UpperBound(long a[], long x) 
			{
				int l=-1, r=a.length;
			    while(l+1<r) 
			    {
			       int m=(l+r)>>>1;
			       if(a[m]<=x) l=m;
			       else r=m;
			    }
			    return l+1;
			}
			
			public static void Sort(long[] a) 
			{
				List<Long> l = new ArrayList<>();
				for (long i : a) l.add(i);
				Collections.sort(l);
				Collections.reverse(l);  //Use to Sort decreasingly 
				for (int i=0; i<a.length; i++) a[i]=l.get(i);
			}
			
			public static void ssort(char[] a)
			{
				List<Character> l = new ArrayList<>();
				for (char i : a) l.add(i);
				Collections.sort(l);
				for (int i=0; i<a.length; i++) a[i]=l.get(i);
			}
		
			
	public static void main(String[] args) throws Exception 
	{
	
		Reader sc = new Reader();
		PrintWriter fout = new PrintWriter(System.out);

		int t = sc.nextInt();
		while(t-- > 0)
		{
			int n = sc.nextInt();
			long[] a = sc.longReadArray(n);
			long ans = 0;
			for(int i = 0;i<n;i++)
			{
				long curr = a[i]-(i+1);
				ans = max(ans,curr);
			}
			fout.println(ans);
		}
		
		fout.close();
	}
}