#include<bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;++i)
using namespace std;

int main() {
    int64_t N, sum = 0;
    cin >> N;
    int A[N];
    rep(i, N) {
        cin >> A[i];
        sum += A[i];
    }
    int64_t oneMove = N * (N + 1) / 2;
    if (sum % oneMove) {
        cout << "NO" << endl;
        return 0;
    }
    int64_t k = sum / oneMove, cnt = 0;
    rep(i, N) {
        int d = A[(i+1)%N] - A[i] - k;
        if (d % N) {
            cout << "NO" << endl;
            return 0;
        }
        cnt += abs(d / N);
    }
    puts(cnt == k ? "YES" : "NO");
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
#define rep(i,n) for(int i=0;i<n;++i)
using namespace std;

int main() {
    int64_t N, sum = 0;
    cin >> N;
    int A[N];
    rep(i, N) {
        cin >> A[i];
        sum += A[i];
    }
    int64_t oneMove = N * (N + 1) / 2;
    if (sum % oneMove) {
        cout << "NO" << endl;
        return 0;
    }
    int64_t k = sum / oneMove, cnt = 0;
    rep(i, N) {
        int d = A[(i+1)%N] - A[i] - k;
        if (d % N) {
            cout << "NO" << endl;
            return 0;
        }
        cnt += abs(d / N);
    }
    puts(cnt == k ? "YES" : "NO");
}
```