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;

signed main () {
	ios_base::sync_with_stdio(false);
	cin.tie(NULL);
	
	int n;
	cin >> n;
	long long s = 0, k = int64_t(n) * int64_t(n + 1) / 2LL;
	vector < int > a(n);
	for (int i = 0; i < n; ++i) {
		cin >> a[i];
		s += a[i];
	}
	if (s % k)
		cout << "NO\n";
	else {
		k = s / k;
		bool fucked_up = 0;
		for (int i = 0; i < n; ++i) {
			int t = k - a[(i + 1) % n] + a[i];
			fucked_up = fucked_up || (t < 0 || t % n);
		}
		if (fucked_up)
			cout << "NO\n";
		else
			cout << "YES\n";
	}
}
