#include <bits/stdc++.h>
using namespace std;
const int mod = 1000000007;
long long logb(long long base, long long x) { return (log(x) / log(base)); }
long long poww(const long long &a, long long b, const long long &m = mod) {
  if (b == 0) return 1;
  long long x = poww(a, b / 2, m);
  x = x * x % m;
  if (b & 1) x = x * a % m;
  return x;
}
void TAK(vector<long long> &v) {
  for (long long i = 0; i < (long long)(v).size(); i++) cin >> v[i];
}
void TAK(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void SET(long long a[], long long n, long long k) {
  for (long long i = 0; i < n; i++) a[i] = k;
}
void PT(vector<long long> const &v) {
  long long n = (long long)(v).size();
  for (long long i = (0); i < (n); i++)
    cout << ("i") << " is " << (i) << " and " << ("v[i]") << " is " << (v[i])
         << "\n";
  ;
  cout << "\n";
}
void PT(long long const a[], long long n) {
  for (long long i = (0); i < (n); i++)
    cout << ("i") << " is " << (i) << " and " << ("a[i]") << " is " << (a[i])
         << "\n";
  ;
  cout << "\n";
}
const long long MAX = 2e5 + 5;
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
long long n = 0, i = 0, j = 0;
void solve() {
  cin >> n;
  vector<long long> v(n);
  TAK(v);
  long long ans = 0;
  for (i = 1; i < n; i++)
    if (v[i] < v[i - 1]) ans += (v[i - 1] - v[i]);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;
const int mod = 1000000007;
long long logb(long long base, long long x) { return (log(x) / log(base)); }
long long poww(const long long &a, long long b, const long long &m = mod) {
  if (b == 0) return 1;
  long long x = poww(a, b / 2, m);
  x = x * x % m;
  if (b & 1) x = x * a % m;
  return x;
}
void TAK(vector<long long> &v) {
  for (long long i = 0; i < (long long)(v).size(); i++) cin >> v[i];
}
void TAK(long long a[], long long n) {
  for (long long i = 0; i < n; i++) cin >> a[i];
}
void SET(long long a[], long long n, long long k) {
  for (long long i = 0; i < n; i++) a[i] = k;
}
void PT(vector<long long> const &v) {
  long long n = (long long)(v).size();
  for (long long i = (0); i < (n); i++)
    cout << ("i") << " is " << (i) << " and " << ("v[i]") << " is " << (v[i])
         << "\n";
  ;
  cout << "\n";
}
void PT(long long const a[], long long n) {
  for (long long i = (0); i < (n); i++)
    cout << ("i") << " is " << (i) << " and " << ("a[i]") << " is " << (a[i])
         << "\n";
  ;
  cout << "\n";
}
const long long MAX = 2e5 + 5;
long long dx[] = {1, 0, -1, 0};
long long dy[] = {0, 1, 0, -1};
long long n = 0, i = 0, j = 0;
void solve() {
  cin >> n;
  vector<long long> v(n);
  TAK(v);
  long long ans = 0;
  for (i = 1; i < n; i++)
    if (v[i] < v[i - 1]) ans += (v[i - 1] - v[i]);
  cout << ans;
}
signed main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  solve();
  return 0;
}

```