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;

long long n, sum, a[100001];

int main () {
	cin >> n;
	for (int i = 1; i <= n; i ++) cin >> a[i], sum += a[i];
	if (sum % (n * (n + 1) / 2) != 0) cout << "NO", exit (0);
	sum /= (n * (n + 1) / 2);
	a[n + 1] = a[1];
	for (int i = 1; i <= n; i ++) {
		if (a[i + 1] - a[i] - sum > 0 || (a[i + 1] - a[i] - sum) % n != 0) cout << "NO", exit (0);
	}
	cout << "YES";
}