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:

 1 #include<stdio.h>
 2 #include<algorithm>
 3 using namespace std;
 4 int main()
 5 {
 6     int n,i,j,ans;
 7     int a[1010];
 8     scanf("%d",&n);
 9     for(i=0; i<n; i++)
10     {
11         scanf("%d",&a[i]);
12     }
13     while(1)
14     {
15         sort(a,a+n);
16         if(a[n-1]==a[0])///互相减，直到都相等
17         {
18             break;
19         }
20         for(j=n-1; j>0; j--)
21         {
22             if(a[j]!=a[j-1])
23             {
24                 a[j]=a[j]-a[j-1];
25             }
26             else
27             {
28                 continue;
29             }
30         }
31     }
32     ans=a[0]*n;
33     printf("%d",ans);
34     return 0;
35 }

找到规律，使用GCD：

 1 #include<stdio.h>
 2 #include<algorithm>
 3 using namespace std;
 4 int gcd(int a,int b)
 5 {
 6     int r;
 7     while(b>0)
 8     {
 9          r=a%b;
10          a=b;
11          b=r;
12     }
13     return a;
14 }
15 int main()
16 {
17     int i,k,n,ans;
18     int a[1010];
19     scanf("%d",&n);
20     for(i=0;i<n;i++)
21     {
22         scanf("%d",&a[i]);
23     }
24     k=a[0];
25     for(i=1;i<n;i++)
26     {
27         ans=gcd(k,a[i]);
28         k=ans;
29     }
30     printf("%d\n",ans*n);
31     return 0;
32 }