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

int main() {
    unsigned long N;
    cin >> N;
    vector<long> a(N);
    for(auto& i : a)cin >> i;

    long t = accumulate(a.begin(), a.end(), 0L);
    auto k = N * (N + 1) / 2;
    if(t % k != 0 || !(t /= k))return 0 & puts("NO");
    vector<long> b(N);
    b[N - 1] = a[N - 1] - a[0] + t;
    for(unsigned long i = 1; i < N; ++i)b[i - 1] = a[i - 1] - a[i] + t;

    for(const auto i : b)if(i < 0 || i % N != 0)return 0 & puts("NO");
    puts("YES");
    return 0;
}
### 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>
using namespace std;

int main() {
    unsigned long N;
    cin >> N;
    vector<long> a(N);
    for(auto& i : a)cin >> i;

    long t = accumulate(a.begin(), a.end(), 0L);
    auto k = N * (N + 1) / 2;
    if(t % k != 0 || !(t /= k))return 0 & puts("NO");
    vector<long> b(N);
    b[N - 1] = a[N - 1] - a[0] + t;
    for(unsigned long i = 1; i < N; ++i)b[i - 1] = a[i - 1] - a[i] + t;

    for(const auto i : b)if(i < 0 || i % N != 0)return 0 & puts("NO");
    puts("YES");
    return 0;
}
```