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;
typedef long long ll;
const int N = 1e5 + 7;

ll a[N];
int main()
{
	int n;
	cin >> n;
	ll sum = 0;
	for (int i=0; i<n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}
	ll x = 1LL * n * (n+1) / 2;
	if (sum % x != 0)
	{
		cout << "NO\n";
		return 0;
	}
	ll y = sum / x;
	for (int i=0; i<n; i++)
		if (a[i] < y)
		{
			cout << "NO\n";
			return 0;
		}
	for (int i=0; i<n; i++)
		if ((a[i] + y) % n != a[(i+1) % n] % n)
		{
			cout << "NO\n";
			return 0;
		}
	cout << "YES\n";
	return 0;
}