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>
#include <vector>
#include <algorithm>
using namespace std;
typedef long long int ll;

bool can(){
	ll n; cin >> n;
	vector<ll> a(n),d(n);
	ll sum=0;
	for(int i=0;i<n;i++){
		cin >> a[i];
		sum+=a[i];
	}
	ll k=n*(n+1)/2;
	if(sum%k){
		return 0;
	}
	k=sum/k;
	for(int i=0;i<n;i++){
		d[i]=a[(i+1)%n]-a[i]-k;
	}
	for(int i=0;i<n;i++){
		if(d[i]%n!=0||d[i]>0)return 0;
	}
	return 1;
}
int main(){
	cin.tie(nullptr);
	ios::sync_with_stdio(false);
	if(can()){
		cout << "YES" << endl; 
	}
	else{
		cout << "NO" << endl;
	}
}