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;
typedef long long LL;
LL MOD = 1000000007;
LL A[100000];
int main() {
	LL N;
	cin >> N;
	LL sum = 0;
	for (int i = 0; i < N; i++) {
		cin >> A[i];
		sum += A[i];
	}
	LL a = N*(N + 1) / 2;
	if (sum%a) {
		cout << "NO" << endl;
		return 0;
	}
	LL k = sum / a;
	for (int i = 0; i < N; i++) {
		LL d = A[(i + 1) % N] - A[i];
		if (d > k || (k - d) % N) {
			cout << "NO" << endl;
			return 0;
		}
	}
	cout << "YES" << endl;
}