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>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long LL;
const LL N = 100005;

LL n,sum = 0,tot,cnt;
LL a[N],b[N];

int main(){
	cin >> n;
	for(LL i = 1;i <= n;i ++) cin >> a[i]; a[0] = a[n];
	for(LL i = 1;i <= n;i ++){
		b[i] = a[i] - a[i - 1];
		sum += a[i];
	}
	if(sum % ((n * (n + 1)) / 2)){ cout << "NO\n"; return 0; }
	tot = sum / ((n * (n + 1)) / 2);
	for(LL i = 1;i <= n;i ++){
		b[i] = b[i] - tot;
		if(b[i] % n || b[i] > 0){ cout << "NO\n"; return 0; }
	}
	cout << "YES\n";
	return 0;
}