#include<iostream>
#include<vector>
#include<numeric>
using namespace std;
typedef long long ll;

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    for(int i = 0; i < n; i++)  cin >> a[i];
    ll sum = accumulate(a.begin(), a.end(), 0ll);
    if(sum%(n*(n+1)/2)){
        cout << "NO" << endl;
        return 0;
    }
    ll k = sum / (n*(n+1)/2);
    for(int i = 0; i < n; i++){
        ll diff = a[i]-a[(i+n-1)%n]-k;
        if(diff > 0 || abs(diff)%n > 0){
            cout << "NO" << endl;
            return 0;
        }
    }
    cout << "YES" << endl;
    return 0;
}
### Prompt

Generate 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<numeric>
using namespace std;
typedef long long ll;

int main(){
    ll n;
    cin >> n;
    vector<ll> a(n);
    for(int i = 0; i < n; i++)  cin >> a[i];
    ll sum = accumulate(a.begin(), a.end(), 0ll);
    if(sum%(n*(n+1)/2)){
        cout << "NO" << endl;
        return 0;
    }
    ll k = sum / (n*(n+1)/2);
    for(int i = 0; i < n; i++){
        ll diff = a[i]-a[(i+n-1)%n]-k;
        if(diff > 0 || abs(diff)%n > 0){
            cout << "NO" << endl;
            return 0;
        }
    }
    cout << "YES" << endl;
    return 0;
}
```