B. Little Elephant and Sorting

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

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.

Sample test(s)

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].

解题说明：题目的要求是通过对一个区间的数字全部加1，实现单调不减数列。可以用贪心的方法来做，既然要单调不减，那么任意两个相邻的数都要实现单调不减。在输入第一个数后，判断下一个数和这个数的差值，这个差值就是下一个数需要增加的数目，依次类推直到最后，对所有差值求和即可。

#include<iostream>
using namespace std;

int main()
{
	int n, i;
	long long a, b, x = 0;
	cin >> n;
	for (i = 1; i <= n; i++)
	{
		b = a;
		cin >> a;
		if (i>1 && a<b)
		{
			x += b - a;
		}
	}
	cout << x<<endl;
	return 0;
}