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

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

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].

題解：記錄之前最高的，若當前比前一個高比之前最高矮，讓之前的先都加到當前這個，若比之前最高的還高，讓之前都增加到之前最高

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define lowbit(x) (x&(-x))
#define INF 0x3f3f3f3f
const int N=110100;
typedef long long ll;
ll n;
ll a[N];
int main()
{
	ll x,l;
    while(scanf("%lld",&n)!=EOF)
    {
        ll ans=0;
        scanf("%d",&a[1]);
        l=a[1];
        for(int i=2;i<=n;i++)
        {
        	scanf("%d",&a[i]);
            if(a[i]<=l&&a[i]>a[i-1])
                ans+=a[i]-a[i-1];
            else if(a[i]>l)
            {
                ans+=l-a[i-1];
                l=a[i];
            }
        }
        if(a[n]<l) ans+=l-a[n];
        cout<<ans<<endl;
    }
	return 0;
}