Codeforces Round #392 (Div. 2) 758A Holiday Of Equality 【按题意模拟】

题目传送门：

A. Holiday Of Equality

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

In Berland it is the holiday of equality. In honor of the holiday the king decided to equalize the welfare of all citizens in Berland by the expense of the state treasury.

Totally in Berland there are n citizens, the welfare of each of them is estimated as the integer ina_i burles (burle is the currency in Berland).

You are the royal treasurer, which needs to count the minimum charges of the kingdom on the king's present. The king can only give money, he hasn't a power to take away them.

Input

The first line contains the integer n (1 ≤ n ≤ 100) — the number of citizens in the kingdom.

The second line contains n integers a₁, a₂, ..., a_n, wherea_i (0 ≤ a_i ≤ 10⁶) — the welfare of thei-th citizen.

Output

In the only line print the integer S — the minimum number of burles which are had to spend.

Examples

Input

5
0 1 2 3 4

Output

10

Input

5
1 1 0 1 1

Output

1

Input

3
1 3 1

Output

4

Input

1
12

Output

0

Note

In the first example if we add to the first citizen 4 burles, to the second3, to the third 2 and to the fourth1, then the welfare of all citizens will equal 4.

In the second example it is enough to give one burle to the third citizen.

In the third example it is necessary to give two burles to the first and the third citizens to make the welfare of citizens equal3.

In the fourth example it is possible to give nothing to everyone because all citizens have12 burles.

题意：问你最少给多少钱能让这些人所拥有的钱数相等?(国王只能给钱，不能往回要钱)

直接排序计算差值的和就好

#include<bits/stdc++.h>
using namespace std;
#define LL long long
#define M(a)  memset(a,0,sizeof(a))
int main()
{
    int ai[105];
    int n;
    while(~scanf("%d",&n))
    {
        for(int i=0;i<n;i++)
        {
            scanf("%d",&ai[i]);
        }
        sort(ai,ai+n);
        int MAX=ai[n-1];
        int sum=0;
        for(int i=0;i<n;i++)
        {
            sum+=(MAX-ai[i]);
        }
        printf("%d\n",sum);
    }
    return 0;
}