Codeforces Round #129 (Div. 2) B. Little Elephant and Sorting

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 thatl ≤ 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

31 2 3

output

0

input

33 2 1

output

2

input

47 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：刚开始以为是要最少的次数把序列排成有序的，后来才发现是要在每个数上加数，然后是数列是非递减序列，我的思路是只要是后面一个数比前面的小，那么就加数直到两个数相等，以此类推，算出总次数。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;int main(){int n,i;long long a,b,sum=0;   while(~scanf("%d",&n))   {       for (i=1;i<=n;i++){b=a; scanf("%d",&a);if (i>1&&a<b){         sum+=b-a;}}cout<<sum<<endl;   }return 0;}