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;
using ll = long long;
using ull = unsigned long long;
#define REP(i,a) for(int i=0;i<(a);i++)
#define MOD 1000000007

int main(){
  ll N;
  cin>>N;
  ll a[N];
  ll sum=0;
  for(int i=0;i<N;i++){
    cin>>a[i];
    sum+=a[i];
  }
  ll x=N*(N+1)/2, k=sum/x;
  bool flag=(sum%x==0);
  for(int i=0;i<N;i++){
    ll d=a[(i+1)%N]-a[i]-k;
    flag&=(d<=0&&d%N==0);
  }
  cout<<(flag?"YES":"NO")<<endl;
  return 0;
}
