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;

vector<long long> a, b;
long long n, sum, k;
int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin >> n;
	a.resize(n);
	b.resize(n);
	for(int i = 0; i < n; i++) cin >> a[i], sum += a[i];
	if(sum % (n*(n+1)/2)) cout << "NO", exit(0);
	k = sum / (n*(n+1)/2);
	for(int i = 0; i < n-1; i++) b[i] = a[i+1] - a[i];
	b[n-1] = a[0] - a[n-1];
	for(int i = 0; i < n; i++) if((k-b[i])%n || k-b[i] < 0) cout << "NO",exit(0);
	cout << "YES";
}
