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 maxn = 1e5 + 10;
int n;
ll tot;
ll a[maxn], b[maxn];
void no() { puts("NO"); exit(0);}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) 
	{
		scanf("%lld", &a[i]);
		tot += a[i];
	}
	ll per = 1ll * n * (n + 1) / 2;
	if(tot % per) no();
	tot /= per;
	a[0] = a[n];
	for(int i = 1; i <= n; ++i) 
	{
		b[i] = tot - (a[i] - a[i - 1]);
		if(b[i] < 0 || b[i] % n) no();
	}
	puts("YES");
	return 0;
}