#include<bits/stdc++.h>
using namespace std;
#define int long long
const int N=100005;
int a[N],n,d[N],sum;
void no(){
	puts("NO");
	exit(0);
}
signed main(){
	scanf("%lld",&n);
	for (int i=1;i<=n;i++)scanf("%lld",&a[i]),sum+=a[i],d[i]=a[i]-a[i-1];
	d[1]=a[1]-a[n];
	int tmp=n*(n+1)/2;
	if (sum%tmp)no();
	int k=sum/tmp,sum2=0;
	for (int i=1;i<=n;i++){
		int x=d[i]+k*(n-1);
		if (x%n!=0)no();
		x/=n;
		if (k-x<0||k-x>k)no();
		sum2+=k-x; 
	}
	printf("%s\n",sum2==k?"YES":"NO");
}
### Prompt

Generate a CPP solution 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<bits/stdc++.h>
using namespace std;
#define int long long
const int N=100005;
int a[N],n,d[N],sum;
void no(){
	puts("NO");
	exit(0);
}
signed main(){
	scanf("%lld",&n);
	for (int i=1;i<=n;i++)scanf("%lld",&a[i]),sum+=a[i],d[i]=a[i]-a[i-1];
	d[1]=a[1]-a[n];
	int tmp=n*(n+1)/2;
	if (sum%tmp)no();
	int k=sum/tmp,sum2=0;
	for (int i=1;i<=n;i++){
		int x=d[i]+k*(n-1);
		if (x%n!=0)no();
		x/=n;
		if (k-x<0||k-x>k)no();
		sum2+=k-x; 
	}
	printf("%s\n",sum2==k?"YES":"NO");
}
```