#include <iostream>
#include <algorithm>

using namespace std;

int main() {
    long n;
    cin >> n;
    long a[n], sum=0;
    for (auto &ai:a) {
        cin >> ai;
        sum += ai;
    }
    long stare = n * (n+1) / 2;
    if (sum % stare) {
        cout << "NO" << '\n';
        return 0;
    }
    long layer = sum / stare,
         prev  = a[n-1];
    for (auto ai:a) {
        long step = prev - ai + layer;
        if (step < 0 || step % n) {
            cout << "NO" << '\n';
            return 0;
        }
        prev = ai;
    }
    cout << "YES" << '\n';
    return 0;
}
### Prompt

Please create a solution in CPP 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 <algorithm>

using namespace std;

int main() {
    long n;
    cin >> n;
    long a[n], sum=0;
    for (auto &ai:a) {
        cin >> ai;
        sum += ai;
    }
    long stare = n * (n+1) / 2;
    if (sum % stare) {
        cout << "NO" << '\n';
        return 0;
    }
    long layer = sum / stare,
         prev  = a[n-1];
    for (auto ai:a) {
        long step = prev - ai + layer;
        if (step < 0 || step % n) {
            cout << "NO" << '\n';
            return 0;
        }
        prev = ai;
    }
    cout << "YES" << '\n';
    return 0;
}
```