#include<stdio.h>
long long a[110000],b[110000];
int main(void)
{
	long long i,j,k=0,n,m,sum=0;
	scanf("%lld",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		sum=sum+a[i];
	}
	if(sum%(n*(n+1)/2)!=0)
	{
		printf("NO\n");
		return 0;
	}
	sum=sum/(n*(n+1)/2);
	b[1]=a[1]-a[n];
	for(i=2;i<=n;i++)
	b[i]=a[i]-a[i-1];
	for(i=1;i<=n;i++)
	{
		if(sum-b[i]<0||(sum-b[i])%n)
		{
			printf("NO\n");
			return 0;
		}
		else
		k=k+(sum-b[i])/n;
	}
	if(k==sum)
	printf("YES\n");
	else
	printf("NO\n");
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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>
long long a[110000],b[110000];
int main(void)
{
	long long i,j,k=0,n,m,sum=0;
	scanf("%lld",&n);
	for(i=1;i<=n;i++)
	{
		scanf("%lld",&a[i]);
		sum=sum+a[i];
	}
	if(sum%(n*(n+1)/2)!=0)
	{
		printf("NO\n");
		return 0;
	}
	sum=sum/(n*(n+1)/2);
	b[1]=a[1]-a[n];
	for(i=2;i<=n;i++)
	b[i]=a[i]-a[i-1];
	for(i=1;i<=n;i++)
	{
		if(sum-b[i]<0||(sum-b[i])%n)
		{
			printf("NO\n");
			return 0;
		}
		else
		k=k+(sum-b[i])/n;
	}
	if(k==sum)
	printf("YES\n");
	else
	printf("NO\n");
	return 0;
}
```