#include <bits/stdc++.h>
using namespace std;
inline void inpint(int &n) {
  n = 0;
  register int ch = getchar_unlocked();
  bool sign = 0;
  while (ch < 48 || ch > 57) {
    if (ch == '-') sign = 1;
    ch = getchar_unlocked();
  }
  while (ch >= 48 && ch <= 57) {
    n = (n << 3) + (n << 1) + ch - 48, ch = getchar_unlocked();
  }
  if (sign) n = -n;
}
inline int sqr(int x) { return x * x; }
inline int cube(int x) { return x * x * x; }
inline long long sqrLL(long long x) { return x * x; }
inline long long cubeLL(long long x) { return x * x * x; }
const long long LLINF = 9223372036854775807LL;
const long long LLINF17 = 100000000000000000LL;
const int INF = 2147483647;
const int INF9 = 1000000000;
const int MOD = 1000000007;
const double eps = 1e-7;
const double PI = acos(-1.0);
int dr[] = {1, 0, -1, 0, -1, 1, 1, -1};
int dc[] = {0, -1, 0, 1, 1, 1, -1, -1};
long long n, arr[100005];
int main() {
  cin >> n;
  for (int(i) = (1); (i) <= (n); (i)++) cin >> arr[i];
  long long lo = 1, ans = 0, cur = 0;
  while (lo < n) {
    arr[lo + 1] += cur;
    if (arr[lo] > arr[lo + 1]) {
      ans += (arr[lo] - arr[lo + 1]);
      cur += arr[lo] - arr[lo + 1];
      arr[lo + 1] = arr[lo];
    }
    lo++;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

The Little Elephant loves sortings.

He has an array a consisting of n integers. Let's number the array elements from 1 to n, then the i-th element will be denoted as ai. The Little Elephant can make one move to choose an arbitrary pair of integers l and r (1 ≤ l ≤ r ≤ n) and increase ai by 1 for all i such that l ≤ i ≤ r.

Help the Little Elephant find the minimum number of moves he needs to convert array a to an arbitrary array sorted in the non-decreasing order. Array a, consisting of n elements, is sorted in the non-decreasing order if for any i (1 ≤ i < n) ai ≤ ai + 1 holds.

Input

The first line contains a single integer n (1 ≤ n ≤ 105) — the size of array a. The next line contains n integers, separated by single spaces — array a (1 ≤ ai ≤ 109). The array elements are listed in the line in the order of their index's increasing.

Output

In a single line print a single integer — the answer to the problem.

Please, do not use the %lld specifier to read or write 64-bit integers in С++. It is preferred to use the cin, cout streams or the %I64d specifier.

Examples

Input

3
1 2 3


Output

0


Input

3
3 2 1


Output

2


Input

4
7 4 1 47


Output

6

Note

In the first sample the array is already sorted in the non-decreasing order, so the answer is 0.

In the second sample you need to perform two operations: first increase numbers from second to third (after that the array will be: [3, 3, 2]), and second increase only the last element (the array will be: [3, 3, 3]).

In the third sample you should make at least 6 steps. The possible sequence of the operations is: (2; 3), (2; 3), (2; 3), (3; 3), (3; 3), (3; 3). After that the array converts to [7, 7, 7, 47].

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
inline void inpint(int &n) {
  n = 0;
  register int ch = getchar_unlocked();
  bool sign = 0;
  while (ch < 48 || ch > 57) {
    if (ch == '-') sign = 1;
    ch = getchar_unlocked();
  }
  while (ch >= 48 && ch <= 57) {
    n = (n << 3) + (n << 1) + ch - 48, ch = getchar_unlocked();
  }
  if (sign) n = -n;
}
inline int sqr(int x) { return x * x; }
inline int cube(int x) { return x * x * x; }
inline long long sqrLL(long long x) { return x * x; }
inline long long cubeLL(long long x) { return x * x * x; }
const long long LLINF = 9223372036854775807LL;
const long long LLINF17 = 100000000000000000LL;
const int INF = 2147483647;
const int INF9 = 1000000000;
const int MOD = 1000000007;
const double eps = 1e-7;
const double PI = acos(-1.0);
int dr[] = {1, 0, -1, 0, -1, 1, 1, -1};
int dc[] = {0, -1, 0, 1, 1, 1, -1, -1};
long long n, arr[100005];
int main() {
  cin >> n;
  for (int(i) = (1); (i) <= (n); (i)++) cin >> arr[i];
  long long lo = 1, ans = 0, cur = 0;
  while (lo < n) {
    arr[lo + 1] += cur;
    if (arr[lo] > arr[lo + 1]) {
      ans += (arr[lo] - arr[lo + 1]);
      cur += arr[lo] - arr[lo + 1];
      arr[lo + 1] = arr[lo];
    }
    lo++;
  }
  cout << ans << endl;
  return 0;
}

```