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>
using ll = long long;

int main(){
	ll N;
	scanf("%lld", &N);
	ll triangle = N*(N + 1) >> 1, sum = 0;
	int a[N];
	for (int i = 0; i < N; ++i){
		scanf("%d", a + i);
		sum += a[i];
	}
	if(sum % triangle){
		printf("NO\n");
		return 0;
	}
	ll sousa = sum/triangle;
	for (int i = 0; i < N; ++i){
		if(sousa + a[i] - a[(i + 1) % N] < 0 || (sousa + a[i] - a[(i + 1) % N]) % N){
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
}