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 <cstdio>
int a[100005];
int main(int argc, char const *argv[])
{
    long long n;
    long long suma =0;
    scanf("%lld",&n);
    for(int i=0;i<n;i++){
        scanf("%d",&a[i]);
        suma += a[i];
    }
    long long sumy = suma/(((1+n)*n)/2);
    if(sumy * (((1+n)*n)/2) != suma ){
        printf("NO\n");
        return 0;
    }


    for (int i=0;i<n-1;i++){
        long long k = a[i+1]-a[i];
        long long y = (sumy - k)/n;
        if(y<0 || y*n != sumy -k){
            printf("NO\n");
            return 0;
        }

    }
    printf("YES\n");


    return 0;
}