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>
long long n, a[100010], k, i, cnt, x, tmp, check;
int main(){
	for(scanf("%lld", &n); i<n; i++){
		scanf("%lld", a+i);
		k+=a[i];
		cnt+=i+1;
	}
	if(k%cnt!=0)	return 0*printf("NO");
	k/=cnt;
	cnt=0;
	for(i=0; i<n; i++){
		tmp=a[i]-a[(i+1)%n];
		if((tmp+k)%n!=0||tmp+k<0)	return 0*printf("NO");
		x=(tmp+k)/n;
		cnt+=x;
		check+=x*(n-i);
	}
	if(cnt!=k||check!=a[0])	return 0*printf("NO");
	puts("YES");
	return 0;
}