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>
typedef long long lint;
int const N=1e5+10;
int n,a[N];
int dif[N];
int main()
{
	scanf("%d",&n);
	if(n==1) {printf("YES"); return 0;}
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	a[n+1]=a[1];
	lint s=(lint)n*(n+1)>>1,sum=0;
	for(int i=1;i<=n;i++) sum+=a[i],dif[i]=a[i+1]-a[i];
	if(sum%s!=0) {printf("NO"); return 0;}
	lint k=sum/s,sumX=0;
	for(int i=1;i<=n;i++)
	{
		lint x=(k-dif[i])/n;
		if(x<0 || x*n!=k-dif[i]) {printf("NO"); return 0;}
		sumX+=x;
	}
	if(sumX==k)	printf("YES");
	else printf("NO");
	return 0;
}