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;
void solve() {
  long long n, ans = 0;
  cin >> n;
  vector<long long> arr(n);
  for (auto &x : arr) cin >> x;
  stack<long long> st;
  for (auto x : arr) {
    if (st.empty())
      st.push(x);
    else {
      if (x == st.top()) continue;
      if (x < st.top())
        st.push(x);
      else if (st.size() == 1) {
        st.pop();
        st.push(x);
      } else {
        long long mn = st.top(), mx = st.top();
        while (st.size() and st.top() <= x) mx = max(mx, st.top()), st.pop();
        mx = st.size() ? min(st.top(), x) : mx;
        st.push(x);
        ans += mx - mn;
      }
    }
  }
  if (!st.empty()) {
    long long mn = st.top(), mx = st.top();
    while (!st.empty()) mx = max(mx, st.top()), st.pop();
    ans += mx - mn;
  }
  cout << ans << endl;
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  cin >> t;
  while (t--) solve();
  return 0;
}
