#include"bits/stdc++.h"
using namespace std;
using ll = long long;

int main() {
    ll N;
    cin >> N;
    vector<ll> A(N);
    for (ll i = 0; i < N; i++) {
        cin >> A[i];
    }

    ll sum = accumulate(A.begin(), A.end(), (ll)0);

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

    ll k = sum / (N * (N + 1) / 2);

    for (ll i = 0; i < N; i++) {
        ll d = A[(i + 1) % N] - A[i] - k;
        if (d > 0 || d % N != 0) {
            cout << "NO" << endl;
            return 0;
        }
    }

    cout << "YES" << endl;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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"
using namespace std;
using ll = long long;

int main() {
    ll N;
    cin >> N;
    vector<ll> A(N);
    for (ll i = 0; i < N; i++) {
        cin >> A[i];
    }

    ll sum = accumulate(A.begin(), A.end(), (ll)0);

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

    ll k = sum / (N * (N + 1) / 2);

    for (ll i = 0; i < N; i++) {
        ll d = A[(i + 1) % N] - A[i] - k;
        if (d > 0 || d % N != 0) {
            cout << "NO" << endl;
            return 0;
        }
    }

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