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>
#include <algorithm>
using namespace std;

const int MAXN = 100003;
int A[MAXN], B[MAXN];

int main() {
	int n;
	scanf("%d", &n);
	long long s = 0;
	for(int i = 1; i <= n; i++) {
		scanf("%d", A + i);
		s += A[i];
		B[i] = A[i] - A[i - 1];
	}
	B[1] = A[1] - A[n];
	bool fg = 1;
	long long t = (long long)(n + 1) * n / 2;
	if(s % t) fg = 0;
	s /= t;
	for(int i = 1; i <= n; i++) {
		B[i] -= s;
		if(B[i] % n || A[i] < s) {
			fg = 0;
			break;
		}
	}
	if(fg) puts("YES");
	else puts("NO");
	return 0;
}
