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;

ll A[101010];
ll n, s, k, c;

void die() { cout << "NO" << endl; exit(0); }

int main()
{
	ios::sync_with_stdio(false);
	cin.tie(NULL);
	
	ll i, t;
	
	cin >> n;
	
	for(i=0; i<n; i++){
		cin >> A[i];
		s += A[i];
	}
	
	k = s / (n * (n + 1) / 2);
	
	if(!k || s % k != 0) die();
	
	for(i=0; i<n; i++){
		t = A[i] + k - A[(i + 1) % n];
		if(t < 0 || t % n != 0) die();
		c += t / n;
	}
	
	if(c != k) die();
	
	cout << "YES" << endl;
	
	return 0;
}