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

int main() {
	int64_t n;
	cin >> n;
	int64_t a[n];
	for (auto &i : a) {
		cin >> i;
	}
	auto sum = accumulate(a, a + n, int64_t(0));
	if (sum % (n * (n + 1) / 2)) {
		cout << "NO" << endl;
		return 0;
	}
	int64_t k = sum / (n * (n + 1) / 2), d[n];
	for (int i = 0; i < n; i++) {
		d[i] = a[(i + 1) % n] - a[i];
	}
	for (auto &i : d) {
		if (k < i || (i - k) % n) {
			cout << "NO" << endl;
			return 0;
		}
	}
	cout << "YES" << endl;
	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() {
	int64_t n;
	cin >> n;
	int64_t a[n];
	for (auto &i : a) {
		cin >> i;
	}
	auto sum = accumulate(a, a + n, int64_t(0));
	if (sum % (n * (n + 1) / 2)) {
		cout << "NO" << endl;
		return 0;
	}
	int64_t k = sum / (n * (n + 1) / 2), d[n];
	for (int i = 0; i < n; i++) {
		d[i] = a[(i + 1) % n] - a[i];
	}
	for (auto &i : d) {
		if (k < i || (i - k) % n) {
			cout << "NO" << endl;
			return 0;
		}
	}
	cout << "YES" << endl;
	return 0;
}
```