#include<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll d[100000], r[100000];
int main()
{
	ll num;
	scanf("%lld", &num);
	ll s = 0;
	for (int i = 0; i < num; i++)scanf("%lld", &d[i]), s += d[i];
	bool f = true;
	if (s % (num*(num + 1) / 2) != 0)f = false;
	s /= (num*(num + 1) / 2);
	ll t = 0;
	for (int i = 0; i < num; i++)
	{
		int p = (i + num - 1) % num;
		if ((d[p] + s - d[i]) % num != 0)f = false;
		r[i] = (d[p] + s - d[i]) / num;
		if (r[i] < 0)f = false;
		t += r[i];
	}
	if (s != t)f = false;
	printf(f ? "YES\n" : "NO\n");
}
### 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<stdio.h>
#include<vector>
#include<algorithm>
using namespace std;
typedef long long ll;
ll d[100000], r[100000];
int main()
{
	ll num;
	scanf("%lld", &num);
	ll s = 0;
	for (int i = 0; i < num; i++)scanf("%lld", &d[i]), s += d[i];
	bool f = true;
	if (s % (num*(num + 1) / 2) != 0)f = false;
	s /= (num*(num + 1) / 2);
	ll t = 0;
	for (int i = 0; i < num; i++)
	{
		int p = (i + num - 1) % num;
		if ((d[p] + s - d[i]) % num != 0)f = false;
		r[i] = (d[p] + s - d[i]) / num;
		if (r[i] < 0)f = false;
		t += r[i];
	}
	if (s != t)f = false;
	printf(f ? "YES\n" : "NO\n");
}
```