Fox and Number Game

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 indexes i and j such that x_i > x_j hold, and then apply assignment x_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 contains nintegers: x₁, x₂, ..., x_n (1 ≤ x_i ≤ 100).

Output

Output a single integer — the required minimal sum.

Examples

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.

题目意思：给你n个数，任意两个数直接，大数可以减去小数得到新的值再赋值给大数，如此反复，直到不出现大数和小数，即所有的数都相等。

解题思路：我看了看数据量很小，于是可以选择使用模拟的方法，将过程模拟了一下，其实能够发现这道题存在着规律，最后所有的数都会变成所有数的最大公约数。

上代码：

模拟法:

#include<stdio.h>
#include<algorithm>
using namespace std;
int main()
{
    int n,i,j,ans;
    int a[1010];
    scanf("%d",&n);
    for(i=0; i<n; i++)
    {
        scanf("%d",&a[i]);
    }
    while(1)
    {
        sort(a,a+n);
        if(a[n-1]==a[0])///互相减，直到都相等
        {
            break;
        }
        for(j=n-1; j>0; j--)
        {
            if(a[j]!=a[j-1])
            {
                a[j]=a[j]-a[j-1];
            }
            else
            {
                continue;
            }
        }
    }
    ans=a[0]*n;
    printf("%d",ans);
    return 0;
}

找到规律，使用GCD：

#include<stdio.h>
#include<algorithm>
using namespace std;
int gcd(int a,int b)
{
    int r;
    while(b>0)
    {
         r=a%b;
         a=b;
         b=r;
    }
    return a;
}
int main()
{
    int i,k,n,ans;
    int a[1010];
    scanf("%d",&n);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
    }
    k=a[0];
    for(i=1;i<n;i++)
    {
        ans=gcd(k,a[i]);
        k=ans;
    }
    printf("%d
",ans*n);
    return 0;
}