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 int ll;
int main(){
	ll n,a[100000],b[100000];
	ll total = 0;
	bool flag = true;
	scanf("%lld",&n);
	for(int i=0;i<n;i++){
		scanf("%lld",a+i);
		total += a[i];
	}
	if(total%(n*(n+1)/2)==0) total /= n*(n+1)/2;
	else flag = false;
	for(int i=0;i<n;i++) if(a[(i+1)%n] - a[i] - total > 0) flag = false;
	for(int i=0;i<n;i++) if((a[(i+1)%n] - a[i] - total)%n!=0) flag = false;
	if(flag) printf("YES\n");
	else printf("NO\n");
}
