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
// be name khoda
#include <bits/stdc++.h>
using namespace std;
typedef long long int ll;
#define pb push_back
const ll maxn=3*1e5+5;
ll a[maxn];
int main()
{
	ll n;cin >> n;
	ll s=0; ll s1=n*(n+1)/2;
	for(ll i=0;i<n;i++)
	{
		cin >> a[i];
		s+=a[i];
	}
	if (n==1)
	{
		cout << "YES";
		return 0;
	}
	if (s%s1!=0)
	{
		cout << "NO";
		return 0;
	}
	s1=s/s1;
	ll s2=s1; ll y=0;
	for(ll i=0;i<n;i++)
	{
		y=a[i]-a[(i+1)%n]+s1;
		if (y<0 || y%n)
		{
			cout << "NO";
			return 0;
		}
		s2-=y/n;
	}
	if (s2==0) cout<<"YES"; else cout<<"NO"; 
}
