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 pair<int, int> P;
typedef long long ll;

int main(){
	ll N, sum = 0; cin >> N;
	ll A[N];
	for(int i=0; i<N; i++){
		cin >> A[i];
		sum += A[i];
	}
	if(sum % (N*(N+1)/2) == 0){
		ll cnt = sum/(N*(N+1)/2), chk = 0;
		for(int i=0; i<N; i++){
			ll gap = A[(i+1)%N] - A[i];
			if((cnt-gap)%N == 0 && 0 <= (cnt-gap)/N && (cnt-gap)/N <= cnt){
				chk += (cnt-gap)/N;
			}else{
				cout << "NO" << "\n";
				return 0;
			}
		}
		if(chk == cnt) cout << "YES" << "\n";
		else cout << "NO" << "\n";
	}else{
		cout << "NO" << "\n";
	}
	return 0;
}