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[];
     scanf("%d",&n);
     for(i=; i<n; i++)
     {
         scanf("%d",&a[i]);
     }
     while()
     {
         sort(a,a+n);
         if(a[n-]==a[])///互相减，直到都相等
         {
             break;
         }
         for(j=n-; j>; j--)
         {
             if(a[j]!=a[j-])
             {
                 a[j]=a[j]-a[j-];
             }
             else
             {
                 continue;
             }
         }
     }
     ans=a[]*n;
     printf("%d",ans);
     return ;
 }

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>)
     {
          r=a%b;
          a=b;
          b=r;
     }
     return a;
 }
 int main()
 {
     int i,k,n,ans;
     int a[];
     scanf("%d",&n);
     for(i=;i<n;i++)
     {
         scanf("%d",&a[i]);
     }
     k=a[];
     for(i=;i<n;i++)
     {
         ans=gcd(k,a[i]);
         k=ans;
     }
     printf("%d\n",ans*n);
     return ;
 }