PROBLEM STATEMENT :

You want to build a temple for snakes. The temple will be built on a mountain range, which can be thought of as n blocks, where height of i-th block is given by hi. The temple will be made on a consecutive section of the blocks and its height should start from 1 and increase by exactly 1 each time till some height and then decrease by exactly 1 each time to height 1, i.e. a consecutive section of 1, 2, 3, .. x-1, x, x-1, x-2, .., 1 can correspond to a temple. Also, heights of all the blocks other than of the temple should have zero height, so that the temple is visible to people who view it from the left side or right side.

You want to construct a temple. For that, you can reduce the heights of some of the blocks. In a single operation, you can reduce the height of a block by 1 unit. Find out minimum number of operations required to build a temple.

INPUT

The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.

The first line of each test case contains an integer n.

The next line contains n integers, where the i-th integer denotes hi

OUTPUT

For each test case, output a new line with an integer corresponding to the answer of that testcase.

CONSTRAINTS

1 ≤ T ≤ 10

2 ≤ N ≤ 105

1 ≤ Hi ≤ 10^9

Example Input

3
3
1 2 1
4
1 1 2 1
5
1 2 6 2 1

Output

0
1
3

MY SOLUTION

#include <stdio.h>
 
void quickSort(int arr[], int low, int high);
int partition (int arr[], int low, int high);
void swap(int* a, int* b);
 
int main(void) {
    // your code goes here
    int t=0, i=0, n=0, move=0, j=0;
    scanf("%d\n", &t);
    for(i=1;i<=t;++i)
    {
        scanf("%d\n", &n);
        int arr[n];
        int i5;
        for(i5=0;i5<n;++i5)
        scanf("%d ", &arr[i5]);
        
        quickSort(arr, 0, n-1);
        
        int max= (arr[n-1]!=arr[n-2])?arr[n-2]+1:arr[n-1];
        move = arr[n-1]-max;
        --max;
        int i3=0;
        for(i3=n-2;i3>0;--max)
        {
            move+=((arr[i3]-max)+(arr[i3-1]-max));
            i3-=2;
            if(max==1)
            {
                break;
            }
        }
        for(j=0;j<=i3;++j)
        {
            move+=arr[j];
        }
        printf("%d\n", move);
    }
    return 0;
}
 
// A utility function to swap two elements
void swap(int* a, int* b)
{
    int t = *a;
    *a = *b;
    *b = t;
}
 
/* This function takes last element as pivot, places
   the pivot element at its correct position in sorted
    array, and places all smaller (smaller than pivot)
   to left of pivot and all greater elements to right
   of pivot */
int partition (int arr[], int low, int high)
{
    int pivot = arr[high];    // pivot
    int i = (low - 1);  // Index of smaller element
    int j;
 
    for (j = low; j <= high- 1; j++)
    {
        // If current element is smaller than or
        // equal to pivot
        if (arr[j] <= pivot)
        {
            i++;    // increment index of smaller element
            swap(&arr[i], &arr[j]);
        }
    }
    swap(&arr[i + 1], &arr[high]);
    return (i + 1);
}
 
/* The main function that implements QuickSort
 arr[] --> Array to be sorted,
  low  --> Starting index,
  high  --> Ending index */
void quickSort(int arr[], int low, int high)
{
    if (low < high)
    {
        /* pi is partitioning index, arr[p] is now
           at right place */
        int pi = partition(arr, low, high);
 
        // Separately sort elements before
        // partition and after partition
        quickSort(arr, low, pi - 1);
        quickSort(arr, pi + 1, high);
    }
}