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>
using namespace std;
typedef long long ll;
ll n,s[100000],tot,N;
int main(void){
    cin>>n;
    for(int i=0;i<n;i++){
        cin>>s[i];
        tot+=s[i];
    }
    N=n*(n+1)/2;
    if(tot%N!=0){
        cout<<"NO"<<endl;
        return 0;
    }
    N=tot/N;
    tot=0;
    for(int i=1;i<n;i++){
        s[i-1]=s[i]-s[i-1]-N;
        if(-s[i-1]%n!=0||s[i-1]>0){
            cout<<"NO"<<endl;
            return 0;
        }
        tot+=-s[i-1]/n;
    }
    if(tot<=N){
        cout<<"YES"<<endl;
    }
    else cout<<"NO"<<endl;
}
