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 LL;
const int N = 1e5 + 5;
LL a[N];
int n;
LL sum, c, cc;
bool solve(){
	c = 1ll*n*(n+1)>>1;
	if(sum % c) return false;
	c = sum / c;
	for(int i = 1; i < n; i ++){
		if(c < a[i] - a[i-1]) return false;
		if((c - a[i] + a[i-1]) % n) return false;
		cc += (c - a[i] + a[i-1]) / n;
	}
	if((c - a[0] + a[n-1]) % n) return false;
	cc += (c - a[0] + a[n-1]) / n;
	if(c == cc) return true;
	return false;
}
int main(){
	scanf("%d", &n);
	for(int i = 0; i < n; i ++){
		scanf("%lld", &a[i]);
		sum += a[i];
	}
	puts(solve()?"YES":"NO");
}