#include<cstdio>
#include<iostream>
long long s[100005];
long long dis[100005];
int main()
{
	long long n,i;
	scanf("%d",&n);
	long long sum=0;
	for(i=1;i<=n;i++)
	scanf("%lld",&s[i]),sum+=s[i];
	if(sum%(n*(n+1)/2))
	{
		printf("NO\n");
		return 0;
	}
	sum/=(n*(n+1))/2;
	dis[1]=s[1]-s[n];
	for(i=2;i<=n;i++)
	dis[i]=s[i]-s[i-1];
	long long cnt=0;
	for(i=1;i<=n;i++)
	{
		if((sum-dis[i]<0)||(sum-dis[i])%n)
		{
			printf("NO\n"); 
			return 0;
		}
		else cnt+=(sum-dis[i])/n;
	}
	if(cnt!=sum)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<cstdio>
#include<iostream>
long long s[100005];
long long dis[100005];
int main()
{
	long long n,i;
	scanf("%d",&n);
	long long sum=0;
	for(i=1;i<=n;i++)
	scanf("%lld",&s[i]),sum+=s[i];
	if(sum%(n*(n+1)/2))
	{
		printf("NO\n");
		return 0;
	}
	sum/=(n*(n+1))/2;
	dis[1]=s[1]-s[n];
	for(i=2;i<=n;i++)
	dis[i]=s[i]-s[i-1];
	long long cnt=0;
	for(i=1;i<=n;i++)
	{
		if((sum-dis[i]<0)||(sum-dis[i])%n)
		{
			printf("NO\n"); 
			return 0;
		}
		else cnt+=(sum-dis[i])/n;
	}
	if(cnt!=sum)printf("NO\n");
	else printf("YES\n");
	return 0;
}
```