#include <bits/stdc++.h>
using namespace std;
const int MAXN = 100005;
typedef long long ll;

ll a[MAXN];

int main()
{
	ll n;
	cin >> n;
	ll sum = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}

	ll hj = (1+n) * n / 2;

	if (sum % hj != 0)
	{
		cout << "NO";
		return 0;
	}

	ll k = sum / hj;

	ll sum_ki = 0;
	for (int i = 0; i < n; i++)
	{
		ll d = a[i] - a[(i-1 + n) % n];
		if ( ((k - d) % n != 0) || (k - d < 0))
		{
			cout << "NO";
			return 0;
		}

		ll ki = (k - d) / n;
		sum_ki += ki;
	}

	if (sum_ki != k) cout << "NO";
	else cout << "YES";
	return 0;
}
### Prompt

Generate a cpp solution to the following problem:

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;
const int MAXN = 100005;
typedef long long ll;

ll a[MAXN];

int main()
{
	ll n;
	cin >> n;
	ll sum = 0;
	for (int i = 0; i < n; i++)
	{
		cin >> a[i];
		sum += a[i];
	}

	ll hj = (1+n) * n / 2;

	if (sum % hj != 0)
	{
		cout << "NO";
		return 0;
	}

	ll k = sum / hj;

	ll sum_ki = 0;
	for (int i = 0; i < n; i++)
	{
		ll d = a[i] - a[(i-1 + n) % n];
		if ( ((k - d) % n != 0) || (k - d < 0))
		{
			cout << "NO";
			return 0;
		}

		ll ki = (k - d) / n;
		sum_ki += ki;
	}

	if (sum_ki != k) cout << "NO";
	else cout << "YES";
	return 0;
}
```