There is a trampoline park with n trampolines in a line. The i-th of which has strength S_i.

Pekora can jump on trampolines in multiple passes. She starts the pass by jumping on any trampoline of her choice. 

If at the moment Pekora jumps on trampoline i, the trampoline will launch her to position i + S_i, and S_i will become equal to max(S_i-1,1). In other words, S_i will decrease by 1, except of the case S_i=1, when S_i will remain equal to 1. 

If there is no trampoline in position i + S_i, then this pass is over. Otherwise, Pekora will continue the pass by jumping from the trampoline at position i + S_i by the same rule as above.

Pekora can't stop jumping during the pass until she lands at the position larger than n (in which there is no trampoline). Poor Pekora!

Pekora is a naughty rabbit and wants to ruin the trampoline park by reducing all S_i to 1. What is the minimum number of passes she needs to reduce all S_i to 1?

Input

The first line contains a single integer t (1 ≤ t ≤ 500) — the number of test cases.

The first line of each test case contains a single integer n (1 ≤ n ≤ 5000) — the number of trampolines.

The second line of each test case contains n integers S_1, S_2, ..., S_n (1 ≤ S_i ≤ 10^9), where S_i is the strength of the i-th trampoline.

It's guaranteed that the sum of n over all test cases doesn't exceed 5000.

Output

For each test case, output a single integer — the minimum number of passes Pekora needs to do to reduce all S_i to 1.

Example

Input


3
7
1 4 2 2 2 2 2
2
2 3
5
1 1 1 1 1


Output


4
3
0

Note

For the first test case, here is an optimal series of passes Pekora can take. (The bolded numbers are the positions that Pekora jumps into during these passes.)

  * [1,4,2,2,2,2,2] 
  * [1,4,1,2,1,2,1] 
  * [1,3,1,2,1,1,1] 
  * [1,2,1,2,1,1,1] 



For the second test case, the optimal series of passes is show below.

  * [2,3] 
  * [1,3] 
  * [1,2] 



For the third test case, all S_i are already equal to 1.
/*

I FOLLOW THE WAY OF THE RIGHT HONOURABLE JIM KWIK,
GOD BLESS YOU SIR

1.  I AM A 'NEWBIE' IN RATING
2.  I WILL FAIL MANY TIMES, BUT I WILL NOT FAIL TO LEARN FROM FAILURE
3.  I WILL NEVER SURRENDER MY GOAL, NO MATTER HOW MUCH I FAIL
4.  I WILL PROVE MY SKILLS WHEN THE TIME COMES
5.  I WILL ALWAYS KEEP TRYING

*/





import java.util.*;
import java.io.*;
public class A {



    static boolean tests =  true;
    static void solve(){
       int n = in.nextInt();
       int[] a = in.ai(n);
       int[] p = new int[n+1];
       int j = 0;
       while (j < n && a[j++] == 1);
       long moves = 0;
       for (int i = j-1; i < n; ++i){
           if (i == n-1){
               moves += Math.max(0, a[i]-1-p[i]);
               break;
           }
           if (a[i] > n-(i+1)){
               int diff = a[i]-(n-(i+1));
               int temp = diff;
               diff = Math.max(0, diff-p[i]);
               p[i] = Math.max(0, p[i]-temp);
               moves += diff;
               a[i] = n-(i+1);
           }
           int uncovered = Math.max(0, a[i]-1-p[i]);
           p[i] = Math.max(0, p[i]-(a[i]-1));
           moves += uncovered;
           p[i+1] += p[i];
           for (int k = 2; k <= a[i]; ++k)
               ++p[i+k];
       }
       out.println(moves);
    }























    static Reader in;
    static PrintWriter out;
    static int int_max = (int)1e9, mod = int_max+7;
    public static void main(String[] args){
        in = new Reader();
        out = new PrintWriter(System.out);
        int t = tests ? in.nextInt() : 1;
        while (t-- > 0) solve();
        out.close();
    }
    static class Reader{
        // Reader credited to SecondThread
        BufferedReader b;
        StringTokenizer s;
        Reader(){
            this.b = new BufferedReader(new InputStreamReader(System.in));
            this.s = new StringTokenizer("");
        }
        String next(){
            while (!s.hasMoreTokens())
                try{s = new StringTokenizer(b.readLine());
                }catch(IOException ignored){}
            return s.nextToken();
        }
        int nextInt(){return Integer.parseInt(next());}
        long nextLong(){return Long.parseLong(next());}
        double nextDouble(){return Double.parseDouble(next());}
        int[] ai(int n){
            int[] a = new int[n];
            for (int i = 0; i < n; ++i)
                a[i] = nextInt();
            return a;
        }
        long[] al(int n){
            long[] a = new long[n];
            for (int i = 0; i < n; ++i)
                a[i] = nextLong();
            return a;
        }
    }
}