CodeForces 205B

 - Little Elephant and Sorting

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit Status Practice CodeForces 205B

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 a_i. The Little Elephant can make one move to choose an arbitrary pair of integers l and r (1 ≤ l ≤ r ≤ n) and increase a_i 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) a_i ≤ a_i + 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 ≤ a_i ≤ 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.

Sample Input

Input

3
1 2 3

Output

0

Input

3
3 2 1

Output

2

Input

4
7 4 1 47

Output

6

Hint

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].

//这题开始以为是要求递减子序列什么的、结果自己在本子上写了下、发现
//其实只要看相邻2项就可以了
#include <iostream>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <queue>
#include <cmath>
using namespace std;
int main()
{
  int n,i,x,y;
  __int64 sum;
  while(scanf("%d",&n)!=EOF)
  {
      sum=0;scanf("%d",&x);
      for(i=1;i<n;i++)
      {
          scanf("%d",&y);
          if(y<x) sum+=x-y;
          x=y;
      }
      printf("%I64d\n",sum);
  }
}