#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath> 

using namespace std;

long long a[100100];

int main()
{
	long long n,sum = 0;
	scanf("%lld",&n);
	for(int i = 1;i <= n;i++)
	{
		scanf("%lld",&a[i]);
		sum += a[i];
	}
	long long t = n * (n + 1) / 2;
	if(sum % t)printf("NO\n");
	else{
		sum = sum / t;
		for(int i = 2;i <= n;i++)
			if(sum + a[i - 1] - a[i] < 0 || (sum + a[i - 1] - a[i]) % n)
			{
				printf("NO\n");
				return 0;
			}
		if(sum + a[n] - a[1] < 0 || (sum + a[n] - a[1]) % n)
		{
			printf("NO\n");
			return 0;
		}
		printf("YES\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 <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath> 

using namespace std;

long long a[100100];

int main()
{
	long long n,sum = 0;
	scanf("%lld",&n);
	for(int i = 1;i <= n;i++)
	{
		scanf("%lld",&a[i]);
		sum += a[i];
	}
	long long t = n * (n + 1) / 2;
	if(sum % t)printf("NO\n");
	else{
		sum = sum / t;
		for(int i = 2;i <= n;i++)
			if(sum + a[i - 1] - a[i] < 0 || (sum + a[i - 1] - a[i]) % n)
			{
				printf("NO\n");
				return 0;
			}
		if(sum + a[n] - a[1] < 0 || (sum + a[n] - a[1]) % n)
		{
			printf("NO\n");
			return 0;
		}
		printf("YES\n");
	}
}
```