L - Fox and Number Game

L - Fox and Number Game

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

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Description

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 iand 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 n integers: x₁, x₂, ..., x_n (1 ≤ x_i ≤ 100).

Output

Output a single integer — the required minimal sum.

Sample Input

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

Hint

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.

簡單的數學題，看網上寫的用的什麼優先排列，其實根本沒必要

ac代碼如下：

#include<stdio.h>
int Euclidean(int a,int b)
{
    if(a<b)
    {
        int t;
        t=a;
        a=b;
        b=t;
    }
    if(a%b==0)
        return b;
    else
    {
        while(a%b!=0)
        {
            int r;
            r=a%b;
            a=b;
            b=r;
        }
        return b;
    }


}//輾轉相除法求最大公因數
int main()
{
    int n,a,b,t;
    while(scanf("%d",&n)!=EOF)
    {
        int k=n;
        scanf("%d",&a);
        n=n-1;
        t=a;
        while(n--)
        {
           scanf("%d",&b);
           t=Euclidean(t,b);
        }
        printf("%d\n",t*k);
    }
    return 0;
}