Codeforces 389A (Greatest common divisor)

Fox and number Game

Time Limit: 1000MS

Memory Limit: 262144KB

64bit IO Format: %i64d &%i64u

Submit Status

Description

Fox Ciel is playing a game with numbers now.

Ciel hasNPositive integers:x_{1,  x 2, ...,   x n . She can do the following operation as many times as Needed:select II 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 do the sum of all numbers as small as possible.}

Ciel to find this minimal sum.

Input

The first line contains an integer n (2≤ n ≤100). Then the second line contains n integers: x_{1, x2, ..., x n (1≤ xi ≤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-on-the-assignment: x _{2 = x2- x 1.}

In the second example the optimal sequence of operations are: x_{3 = x3- x 2, x2 = x2- x1.}

Source

Codeforces Round #228 (Div. 2) The problem: to convert the title, is to find the N number of greatest common divisor multiplied by N.

#include <iostream>#include<stdlib.h>#include<stdio.h>#include<algorithm>#include<string.h>#include<math.h>using namespaceStd;typedefLong Longll;ll gcd (ll a,ll b) {if(b==0)returnA; Else returnGCD (b,a%b);}intMain () {inti,n,ans,a[ the];  while(cin>>N) {cin>>a[0]; Ans=a[0];  for(i=1; i<n;i++) {cin>>A[i]; Ans=gcd (Ans,a[i]); } cout<<ans*n<<Endl; }    return 0;}

Codeforces 389A (Greatest common divisor)