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;
typedef long long ll;
ll num[100005]={0};
int main(){
  ll n,sum=0;
  cin>>n;
  for(int i=0;i<n;i++){
    cin>>num[i];
    sum+=num[i];
  }
  ll kn=n*(n+1)/2;
  if(sum%kn!=0){
    cout<<"NO"<<endl;
    return 0;
  }
  ll kai=sum/kn;
  for(int i=0;i<n;i++){
    int nxt=(i+1)%n;
    ll del=num[nxt]-num[i]-kai;
    if(del>0){
      cout<<"NO"<<endl;
      return 0;
    }
    del*=-1;
    if(del%n!=0){
      cout<<"NO"<<endl;
      return 0;
    }
  }
  cout<<"YES"<<endl;
  return 0;
}
