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 a[N], n, b[N];
long long sum, sum2;
long long cak;

int main() {
	cin >> n;
	for(int i = 0; i < n; ++i) cin >> a[i], sum += a[i];
	sum2 = 1LL * n * (n + 1) / 2;
	if(sum % sum2 != 0) return cout << "NO", 0;
	sum /= sum2;
	for(int i = 0; i < n; ++i)
		if((sum + a[i] - a[(i + 1) % n]) % n != 0) return cout << "NO", 0;
		else if(sum + a[i] - a[(i + 1) % n] < 0) return cout << "NO", 0;
		else cak += (sum + a[i] - a[(i + 1) % n]) / n;
	if(cak != sum) return cout << "NO", 0;
	cout << "YES";
}