#include <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <map>
#include <queue>
#define rep(i,n) for(int i = 0; i < n; i++)
#define INF 1000000000
#define MOD 1000000007
#define ll long long
using namespace std;
int n;
ll sum,ans;
ll a[100000];
ll sa[100000];
int main(){
    cin>>n;
    rep(i,n){cin>>a[i]; sum+=a[i];}
    ll set=n*(n+1)/2;
    ll kai=sum/set;
    rep(i,n){
        sa[i]=kai-a[(i+1)%n]+a[i];
        ans+=(sa[i]/n)*(n-i);
    }
    if (ans==a[0]){
        cout<<"YES"<<endl;
    }else{
        cout<<"NO"<<endl;
    }
    return 0;
}
### Prompt

Your task is to create a CPP solution to 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 <iostream>
#include <vector>
#include <cstdio>
#include <algorithm>
#include <map>
#include <queue>
#define rep(i,n) for(int i = 0; i < n; i++)
#define INF 1000000000
#define MOD 1000000007
#define ll long long
using namespace std;
int n;
ll sum,ans;
ll a[100000];
ll sa[100000];
int main(){
    cin>>n;
    rep(i,n){cin>>a[i]; sum+=a[i];}
    ll set=n*(n+1)/2;
    ll kai=sum/set;
    rep(i,n){
        sa[i]=kai-a[(i+1)%n]+a[i];
        ans+=(sa[i]/n)*(n-i);
    }
    if (ans==a[0]){
        cout<<"YES"<<endl;
    }else{
        cout<<"NO"<<endl;
    }
    return 0;
}
```