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;
#define int long long
int mod=1e9+7;

signed main(){
  int n;
  cin>>n;
  vector<int> a(n);
  int sum=0;
  for(int i=0;i<n;i++){
    cin>>a[i];
    sum+=a[i];
  }
  int ka=n*(n+1)/2;
  if(sum%ka!=0){
    cout<<"NO"<<endl;
    return 0;
  }
  int x=sum/ka;
  vector<int> b(n);
  for(int i=0;i<n-1;i++){
    b[i]=a[i]-a[i+1]+x;
  }
  b[n-1]=a[n-1]-a[0]+x;
  bool f=true;
  for(int i=0;i<n;i++){
    //cout<<b[i]<<endl;
    if(b[i]<0||b[i]%n!=0)f=false;
  }
  if(f)cout<<"YES"<<endl;
  else cout<<"NO"<<endl;

}
