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 <iostream>
using namespace std;

typedef long long ll;
ll N;
ll A[100010];
ll B[100010];

int main(){
	cin >> N;
	ll sum = 0;
	for(int i=0;i<N;i++){
		cin >> A[i];
		sum += A[i];
	}
	ll t = (N*(N+1))/2;
	if(sum%t!=0) cout << "NO" << endl;
	else{
		for(int i=0;i<N;i++){
			B[i] = A[(i+1)%N]-A[i]-sum/t;
		}
		bool judge = true;
		for(int i=0;i<N;i++){
			if(B[i]>0 || B[i]%N!=0) judge = false;
		}
		if(judge) cout << "YES" << endl;
		else cout << "NO" << endl;
	}
}