#include<stdio.h>
long long a[100010],b[100010];
int main()
{
	long long n,i,sum,t,t1,t2,p;
	while(scanf("%lld",&n)!=EOF)
	{
		sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%lld",&a[i]);
			sum=sum+a[i];
		}
		t1=n*(n+1)/2;
		if(sum%t1!=0)
		{
			printf("NO\n");
			continue;
		}
		t2=sum/t1;
		for(i=1;i<n;i++)
		b[i]=a[i+1]-a[i];
		b[n]=a[1]-a[n];
		p=0;
		for(i=1;i<=n;i++)
		{
			t=t2-b[i];
			if(t<0)
			break;
			if(t%n!=0)
			break;
			p=p+t/n;
		}
		if(i!=n+1||p!=t2)
		printf("NO\n");
		else
		printf("YES\n");
	}
	return 0;
}
### Prompt

Please create a solution in CPP to 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<stdio.h>
long long a[100010],b[100010];
int main()
{
	long long n,i,sum,t,t1,t2,p;
	while(scanf("%lld",&n)!=EOF)
	{
		sum=0;
		for(i=1;i<=n;i++)
		{
			scanf("%lld",&a[i]);
			sum=sum+a[i];
		}
		t1=n*(n+1)/2;
		if(sum%t1!=0)
		{
			printf("NO\n");
			continue;
		}
		t2=sum/t1;
		for(i=1;i<n;i++)
		b[i]=a[i+1]-a[i];
		b[n]=a[1]-a[n];
		p=0;
		for(i=1;i<=n;i++)
		{
			t=t2-b[i];
			if(t<0)
			break;
			if(t%n!=0)
			break;
			p=p+t/n;
		}
		if(i!=n+1||p!=t2)
		printf("NO\n");
		else
		printf("YES\n");
	}
	return 0;
}
```