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 <bits/stdc++.h>

using namespace std;

const int N = 1000100;

int a[N];
int d[N];

int main() {
	int n;
	cin >> n;
	long long v = (n*1ll*(n+1))/2;
	long long sum = 0;
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
		sum += a[i];
	}
	if(sum % v == 0) {
		sum/=v;
		for (int i = 0; i < n; ++i) {
			d[i] = a[i] - a[(i+1)%n];
		}

		for (int i = 0; i < n; ++i) {
			d[i] += sum;
		}
		int ok = 1;
		for (int i =0 ; i < n; ++i) {
			if(!(d[i]%n == 0 && d[i] >= 0)) {
				ok = 0;
			}
		}
		if (ok) puts("YES");
		else puts("NO");
	} else {

		puts("NO");

	}

}
