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>
#include <algorithm>
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define rep1(i,n) for(int i=1;i<=(int)(n);i++)

using namespace std;
typedef long long int lli;
const int MAX=1e5;

int main(){
	lli N;
    cin>>N;
	lli A[MAX+1];
	rep1(i,N) cin>>A[i];

	bool ans=true;
	lli sum=0;
	rep1(i,N) sum+=A[i];
	
	if(sum%(N*(N+1)/2)!=0) ans=false;
	else{
		lli t=sum/(N*(N+1)/2);
		rep1(i,N)A[i]-=i*t;
		rep1(i,N-1)if(A[i]<A[i+1]||(A[i]-A[i+1])%N!=0) ans=false;
	}
	
	if(ans)cout<<"YES";
	else cout<<"NO";
}
