Codeforces Round #129 (Div. 2) B

Description

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 %I64dspecifier.

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].

题意：可以把区间的数增加相同的值（值的范围是1-n），让数列成为非递减

解法：当然让不符合要求的数字增加到最近的最大值，比如 7 4 1变成7 7 7 ，我们只需要计算7 4,4 1之间的差值就行（增加同一个数，两个数的差值不变的）

#include<bits/stdc++.h>
using namespace std;
long long a[100005];
int pos;
int n;
int d;
int main()
{
    long long sum=0;
    int pos=0;
    int flag=0;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>a[i];
    }
    d=a[1];
    for(int i=2;i<=n;i++)
    {
       if(a[i]<a[i-1])
       {
           sum+=(a[i-1]-a[i]);
         //  pos=(d-a[i]);
       }
     //  cout<<a[i]<<"A"<<endl;
    }
    cout<<sum<<endl;
    return 0;
}