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 = 1e5 + 5;

int n;
long long a[N];

int main() {
	ios_base::sync_with_stdio(false);
	cin >> n;
	for (int i = 1; i <= n; ++i) cin >> a[i];
	long long sum = 0;
	for (int i = 1; i <= n; ++i) sum += a[i];
	long long tot = 1LL * n * (n + 1) / 2LL;
	if (sum % tot) return cout << "NO\n",0;
	tot = sum / tot;
	for (int i = 1; i <= n; ++i) {
		long long d = a[i] - (i == 1 ? a[n] : a[i - 1]);
		d -= tot;
		if (d > 0) return cout << "NO\n",0;
		d = -d;
		if (d % n) return cout << "NO\n",0;
	}
	cout << "YES\n";
}