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 <stdio.h>
#include <iostream>
#include <vector>
#define ll long long
#define rep(i,n) for(ll i=0;i<n;i++)
#define vecll vector<ll>
using namespace std;
int in() {int x;scanf("%d",&x);return x;}
bool f(ll n){
    ll sum=0,cnt=0,p=n*(n+1)/2;
    vecll a(n);
    rep(i,n) sum+=a[i]=in();
    if(sum%p!=0)return false;
    ll num=sum/p;
    rep(i,n){
        ll k=a[(i+1)%n]-a[i]-num;
        if(k>0||k%n!=0)return false;
        cnt-=k/n;
    }
    return cnt==num;
}
int main(){
    string a=f(in())?"YES":"NO";
    cout<<a;
}