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>
#define N 100005
using namespace std;
typedef long long ll;
ll n,a[N],sum;

int main(){
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>a[i];
    sum+=a[i];
  }
  ll num=n*(n+1)/2;
  if(!(sum%num)){
    ll cnt=sum/num,ans=0;
    bool f=true;
    for(int i=0;i<n;i++){
      ll cal=a[(i+1)%n]-a[i];
      if((cnt-cal)%n)f=false;
      cal=(cnt-cal)/n;
      if(cal<0)f=false;
      else ans+=cal;
    }
    if(f&&ans==cnt)cout<<"YES"<<endl;
    else cout<<"NO"<<endl;
  }else cout<<"NO"<<endl;
  return 0;
}
