#include <bits/stdc++.h>
typedef long long i64;
using std::cout;
using std::endl;
using std::cin;
 
int main(){
    i64 n, sum = 0; cin >> n;
    std::vector<int> a(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
    }

    if(sum % (n * (n + 1) / 2)) {
        cout << "NO" << endl;
        return 0;
    }
    sum /= n * (n + 1) / 2;

    for(int i = 0; i < n; i++) {
        i64 d = a[(i + 1) % n] - a[i] - sum;
        if(d % n or d > 0) {
            cout << "NO" << endl;
            return 0;
        }
    }

    cout << "YES" << endl;
    return 0;
}
### 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 i64;
using std::cout;
using std::endl;
using std::cin;
 
int main(){
    i64 n, sum = 0; cin >> n;
    std::vector<int> a(n);
    for(int i = 0; i < n; i++) {
        cin >> a[i];
        sum += a[i];
    }

    if(sum % (n * (n + 1) / 2)) {
        cout << "NO" << endl;
        return 0;
    }
    sum /= n * (n + 1) / 2;

    for(int i = 0; i < n; i++) {
        i64 d = a[(i + 1) % n] - a[i] - sum;
        if(d % n or d > 0) {
            cout << "NO" << endl;
            return 0;
        }
    }

    cout << "YES" << endl;
    return 0;
}
```