#include <iostream>
using namespace std;
int arr[100005];
int main()
{
	int n;
	scanf("%d",&n);
	long long sum=0,cnt=0;
	for (int i=0;i<n;i++)
	scanf("%d",&arr[i]),sum+=arr[i];
	long long tmp=1LL*n*(n+1)/2;
	if (sum%tmp)
	{
		printf("NO");
		return 0;
	}
	sum/=tmp;
	for (int i=0;i<n;i++)
	{
		long long cur=arr[i]-(arr[(i+1)%n]-sum);
		if (cur<0 || cur%n)
		{
			printf("NO");
			return 0;
		}
		cnt+=cur/n;
	}
	printf((cnt==sum)? "YES":"NO");
}
### Prompt

Develop a solution in Cpp 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 <iostream>
using namespace std;
int arr[100005];
int main()
{
	int n;
	scanf("%d",&n);
	long long sum=0,cnt=0;
	for (int i=0;i<n;i++)
	scanf("%d",&arr[i]),sum+=arr[i];
	long long tmp=1LL*n*(n+1)/2;
	if (sum%tmp)
	{
		printf("NO");
		return 0;
	}
	sum/=tmp;
	for (int i=0;i<n;i++)
	{
		long long cur=arr[i]-(arr[(i+1)%n]-sum);
		if (cur<0 || cur%n)
		{
			printf("NO");
			return 0;
		}
		cnt+=cur/n;
	}
	printf((cnt==sum)? "YES":"NO");
}
```