<codeforces>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 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 nintegers,
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 thecin, cout streams
or the %I64d specifier.

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].

#include <iostream>
using namespace std;
int main()
{
int n,a,b;
long long int add;
while(cin>>n>>a)
{
add=0;
while(--n)
{
cin>>b;
if(a>b)
{
add+=a-b;
}
a=b;
}
cout<<add<<endl;
}
return 0;
}