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;

bool solve() {
	long long n;
	cin >> n;

	vector<long long> a(n);
	long long sum = 0;
	for (int i = 0; i < n; i++) {
		scanf("%d", &a[i]);
		sum += a[i];
	}

	long long ss = n * (n + 1) / 2;
	if (sum % ss != 0) return false;

	vector<long long> d(n);
	long long cnt = 0;
	for (int i = 0; i < n; i++) {
		d[i] = (a[(i + 1) % n] - a[i]);
		d[i] -= sum / ss;
		if (d[i] > 0) return false;
		if (d[i] % n == 0) cnt += -d[i] / n;
	}

	return cnt == sum / ss;
}

int main() {
	puts(solve() ? "YES" : "NO");
}
