There are N boxes arranged in a circle. The i-th box contains A_i stones.

Determine whether it is possible to remove all the stones from the boxes by repeatedly performing the following operation:

* Select one box. Let the box be the i-th box. Then, for each j from 1 through N, remove exactly j stones from the (i+j)-th box. Here, the (N+k)-th box is identified with the k-th box.



Note that the operation cannot be performed if there is a box that does not contain enough number of stones to be removed.

Constraints

* 1 ≦ N ≦ 10^5
* 1 ≦ A_i ≦ 10^9

Input

The input is given from Standard Input in the following format:


N
A_1 A_2 … A_N


Output

If it is possible to remove all the stones from the boxes, print `YES`. Otherwise, print `NO`.

Examples

Input

5
4 5 1 2 3


Output

YES


Input

5
6 9 12 10 8


Output

YES


Input

4
1 2 3 1


Output

NO
#include <stdio.h>
int a[100010];
int main(){
    int i,n;
    long long s=0,t=0,m;
    scanf("%d",&n);
    for(i=1;i<=n;++i)scanf("%d",a+i),s+=a[i];
    m=(long long)n*(n+1)/2;
    if(s%m){puts("NO");return 0;}
    m=s/m;
    a[n+1]=a[1];
    for(i=1;i<=n;++i){
        s=a[i+1]-a[i];
        s+=m*(n-1);
        if(s%n||m<s/n){puts("NO");return 0;}
        t+=m-s/n;
    }
    if(t!=m)puts("NO");
    else puts("YES");
    return 0;
}
