【ECJTU_ACM 11级队员2012年暑假训练赛（7） E Little Elephant and Sorting】

E - 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 isuch 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 %I64dspecifier.

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].

#include <iostream>
#include <stdio.h>
using namespace std;

typedef unsigned long long int longint;

#define MAXN 100100

longint iMap[MAXN];
longint n;

void iInit()
{
    for (longint i = 0; i < n; i++)
    {
        cin >> iMap[i];
    }
}

void iCalAnswer()
{
    longint iCount = 0;
    for (longint i = 1; i < n; i++)
    {
        if (iMap[i] < iMap[i-1])
        {
            iCount += (iMap[i-1] - iMap[i]);
        }
    }
    cout << iCount << endl;
}

int main()
{
    //freopen("in.dat", "r", stdin);
    while (cin >> n)
    {
        iInit();
        iCalAnswer();
    }
    return 0;
}

// end 
// ism