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<string>
using namespace std;

#define FOR(k,m,n) for(int (k)=(m);(k)<(n);(k)++)
#define rep(i,n) FOR((i),0,(n))

typedef long long ll;

const int INF=1e9+7;
const int MAX_N=1e5+5;

ll A[MAX_N];
int main(){
	ll n;
	cin>>n;
	rep(i,n)cin>>A[i];
	ll sum=0;
	rep(i,n)sum+=A[i];
	if(((2*sum)%(n*(n+1)))!=0){
		cout<<"NO"<<endl;
		return 0;
	}
	ll k=2*sum/(n*(n+1));
	ll a=0;
	rep(i,n){
		if((A[i]-A[(i+1)%n]+k)>=0 && (A[i]-A[(i+1)%n]+k)%n==0)a+=((A[i]-A[(i+1)%n]+k)/n);
	}
	if(a==k){
		cout<<"YES"<<endl;
	}
	else cout<<"NO"<<endl;
}
