Omkar is building a waterslide in his water park, and he needs your help to ensure that he does it as efficiently as possible.

Omkar currently has n supports arranged in a line, the i-th of which has height a_i. Omkar wants to build his waterslide from the right to the left, so his supports must be nondecreasing in height in order to support the waterslide. In 1 operation, Omkar can do the following: take any contiguous subsegment of supports which is nondecreasing by heights and add 1 to each of their heights. 

Help Omkar find the minimum number of operations he needs to perform to make his supports able to support his waterslide!

An array b is a subsegment of an array c if b can be obtained from c by deletion of several (possibly zero or all) elements from the beginning and several (possibly zero or all) elements from the end.

An array b_1, b_2, ..., b_n is called nondecreasing if b_i≤ b_{i+1} for every i from 1 to n-1.

Input

Each test contains multiple test cases. The first line contains the number of test cases t (1 ≤ t ≤ 100). Description of the test cases follows.

The first line of each test case contains an integer n (1 ≤ n ≤ 2 ⋅ 10^5) — the number of supports Omkar has.

The second line of each test case contains n integers a_{1},a_{2},...,a_{n} (0 ≤ a_{i} ≤ 10^9) — the heights of the supports.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, output a single integer — the minimum number of operations Omkar needs to perform to make his supports able to support his waterslide.

Example

Input


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


Output


3
2
0

Note

The subarray with which Omkar performs the operation is bolded.

In the first test case:

  * First operation:

[5, 3, 2, 5] → [5, 3, 3, 5]

  * Second operation:

[5, 3, 3, 5] → [5, 4, 4, 5]

  * Third operation:

[5, 4, 4, 5] → [5, 5, 5, 5]




In the third test case, the array is already nondecreasing, so Omkar does 0 operations.
#include <bits/stdc++.h>
using namespace std;
long long pf[10000001];
int gcd(int a, int b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
void prime() {
  memset(pf, 0, 10000001);
  pf[0] = pf[1] = 1;
  for (long long i = 2; i < 10000001; i++) {
    if (pf[i] == 0) {
      for (long long j = i * i; j < 10000001; j = j + i) pf[j] = 1;
    }
  }
}
int lcm(int a, int b) { return a * b / gcd(a, b); }
int check_ps(long long n) {
  double sqrt_n = sqrt(n);
  if (sqrt_n == int(sqrt_n)) {
    return 1;
  } else {
    return 0;
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t;
  cin >> t;
  while (t--) {
    long long n;
    cin >> n;
    long long a[n];
    long long maxi = 0, maxp = -1;
    for (int i = 0; i < n; i++) {
      cin >> a[i];
    }
    long long l = 0, j = -1, d = 0, mini = a[0], ans = 0;
    for (int i = 1; i < n; i++) {
      if (a[i] < a[i - 1]) {
        ans += a[i - 1] - a[i];
        d = 0;
      }
    }
    cout << ans << endl;
  }
  return 0;
}
