#include<bits/stdc++.h>
using namespace std;
typedef long long ll;

const int MAX_N = 1e5;
ll N;
ll a[MAX_N+5];

int main(){
    cin >> N;
    for(int i=0;i<N;i++) cin >> a[i];

    bool flg = true;

    ll sum = 0, p = N * (N+1) / 2, sy = 0, m;
    for(int i=0;i<N;i++) sum += a[i];

    if (sum%p) flg = false;

    m = sum / p;
    for(int i=1;i<=N;i++){
        ll d = a[i%N] - a[i-1];

        if((m-d)%N || m < d) flg = false;
        sy += (m-d) / N;
    }

    if(flg && sy==m) cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}
### 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;
typedef long long ll;

const int MAX_N = 1e5;
ll N;
ll a[MAX_N+5];

int main(){
    cin >> N;
    for(int i=0;i<N;i++) cin >> a[i];

    bool flg = true;

    ll sum = 0, p = N * (N+1) / 2, sy = 0, m;
    for(int i=0;i<N;i++) sum += a[i];

    if (sum%p) flg = false;

    m = sum / p;
    for(int i=1;i<=N;i++){
        ll d = a[i%N] - a[i-1];

        if((m-d)%N || m < d) flg = false;
        sy += (m-d) / N;
    }

    if(flg && sy==m) cout << "YES" << endl;
    else cout << "NO" << endl;
    return 0;
}
```