#include <iostream>
#include <vector>
#include <cstdint>

using namespace std;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	int64_t n;
	cin >> n;

	vector<int64_t> numbers(n);

	int64_t sum = 0;
	for (int64_t& x : numbers)
	{
		cin >> x;

		sum += x;
	}

	int64_t boxsum = n*(n+1)/2;
	int64_t boxcube = n*boxsum;

	bool ok = true;
	for (int64_t i=0 ; i<n ; i++)
	{
		int64_t chk = sum + boxsum * (numbers[i] - numbers[(i+1)%n]);
		if (chk<0 || (chk%boxcube))
		{
			ok = false;
		}
	}

	if (ok) cout << "YES" << endl;
	else cout << "NO" << endl;
}

### Prompt

Create a solution in cpp for 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 <iostream>
#include <vector>
#include <cstdint>

using namespace std;

int main()
{
	ios_base::sync_with_stdio(false);
	cin.tie(nullptr);

	int64_t n;
	cin >> n;

	vector<int64_t> numbers(n);

	int64_t sum = 0;
	for (int64_t& x : numbers)
	{
		cin >> x;

		sum += x;
	}

	int64_t boxsum = n*(n+1)/2;
	int64_t boxcube = n*boxsum;

	bool ok = true;
	for (int64_t i=0 ; i<n ; i++)
	{
		int64_t chk = sum + boxsum * (numbers[i] - numbers[(i+1)%n]);
		if (chk<0 || (chk%boxcube))
		{
			ok = false;
		}
	}

	if (ok) cout << "YES" << endl;
	else cout << "NO" << endl;
}

```