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<cstdio>
#include<iostream>
using namespace std;
typedef long long LL;
const int N = 100009;
LL g[N], cnt, v, vn;
LL n;
int main()
{
	std::ios::sync_with_stdio(false);
	cin >> n;
	for(LL i = 1; i <= n; i++)
		cin >> g[i], cnt += g[i];
	vn = n * (n + 1) / 2;
	if(cnt % vn)
	{
		puts("NO");
		return 0;
	}
	LL p = g[1];
	for(LL i = 1; i < n; i++)
		g[i] = g[i+1] - g[i];
	g[n] = p - g[n];
	LL d = cnt / vn;
	for(LL i = 1; i <= n; i++)
	{
		g[i] -= d;
		if(g[i] % n || g[i] > 0)
		{
			puts("NO");
			return 0;
		}
	}
	puts("YES");
	return 0;
}