#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll N=1e5+5;
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n;cin>>n;
    ll a[n],sum=0;
    for(ll i=0;i<n;i++)cin>>a[i],sum+=a[i];
    ll tot = (n*(n+1))/2;
    if(sum%tot)return cout<<"NO" , 0;
    ll k = sum/tot;
    //cout<<"K: "<<k<<endl;
    for(ll i=0;i<n;i++){
        ll diff = a[(i+1)%n] - a[i];
        ll val = k - diff;
        if(val>=0 && val%n==0)continue;
        return cout<<"NO" , 0;
    }
    cout<<"YES";
}

### Prompt

In CPP, your task is to solve the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
typedef long long ll;
using namespace std;
const ll N=1e5+5;
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n;cin>>n;
    ll a[n],sum=0;
    for(ll i=0;i<n;i++)cin>>a[i],sum+=a[i];
    ll tot = (n*(n+1))/2;
    if(sum%tot)return cout<<"NO" , 0;
    ll k = sum/tot;
    //cout<<"K: "<<k<<endl;
    for(ll i=0;i<n;i++){
        ll diff = a[(i+1)%n] - a[i];
        ll val = k - diff;
        if(val>=0 && val%n==0)continue;
        return cout<<"NO" , 0;
    }
    cout<<"YES";
}

```