Fox and Number Game

A. Fox and Number Game

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Fox Ciel is playing a game with numbers now.

Ciel has n positive integers: x₁,x₂, ...,x_n. She can do the following operation as many times as needed: select two different indexesi and j such thatx_i >x_j hold, and then apply assignmentx_i =x_i -x_j. The goal is to make the sum of all numbers as small as possible.

Please help Ciel to find this minimal sum.

Input

The first line contains an integer n (2 ≤ n ≤ 100). Then the second line containsn integers: x₁,x₂, ...,x_n (1 ≤ x_i ≤ 100).

Output

Output a single integer — the required minimal sum.

Sample test(s)

Input

2
1 2

Output

2

Input

3
2 4 6

Output

6

Input

2
12 18

Output

12

Input

5
45 12 27 30 18

Output

15

Note

In the first example the optimal way is to do the assignment: x₂ =x₂ -x₁.

In the second example the optimal sequence of operations is: x₃ =x₃ -x₂,x₂ =x₂ -x₁.

 

思路：这个题的最后结果一定是n个相同的数，所以在n个数不相等的情况下进行循环，让每一个数减去比他小的。。

 

#include <stdio.h>
#include <stdlib.h>
int main()
{
    int n,m,i,j;
    int a[100];
    while(scanf("%d",&n)!=EOF)
    {
        for(i=0; i<n; i++)
        {
            scanf("%d",&a[i]);
        }
        int s=0;
        for(i=0; i<n-1; i++)
        {
            if(a[i]==a[i+1])
                ++s;
        }
        while(s!=n-1)
        {
            for(i=0; i<n; i++)
            {
                for(j=0; j<n; j++)
                {
                    if(i!=j)
                    {
                        if(a[i]>a[j])
                            a[i]=a[i]-a[j];
                    }
                }
            }
            s=0;
            for(i=0; i<n-1; i++)
            {
                if(a[i]==a[i+1])
                    ++s;
            }
        }
        printf("%d\n",n*a[0]);
    }
    return 0;
}