#include<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define LL long long
#define N 110000
using namespace std;
int n;
LL a[N],d[N],k;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {scanf("%lld",&a[i]);k+=a[i];}
	LL t=1ll*n*(n+1)/2;
	if(k%t) {printf("NO\n");return 0;}
	k/=t;
	a[0]=a[n];
	for(int i=1;i<=n;i++) d[i]=a[i]-a[i-1];
	t=0;
	for(int i=1;i<=n;i++)
	{
		LL tmp=k-d[i];
		if(tmp<0 || tmp%n) {printf("NO\n");return 0;}
		t+=tmp/n;
	}
	if(t!=k) printf("NO\n");
	else printf("YES\n");
	return 0;
}
### Prompt

Create a solution in Cpp for 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<cstring>
#include<cstdlib>
#include<cstdio>
#include<cmath>
#include<iostream>
#define LL long long
#define N 110000
using namespace std;
int n;
LL a[N],d[N],k;
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) {scanf("%lld",&a[i]);k+=a[i];}
	LL t=1ll*n*(n+1)/2;
	if(k%t) {printf("NO\n");return 0;}
	k/=t;
	a[0]=a[n];
	for(int i=1;i<=n;i++) d[i]=a[i]-a[i-1];
	t=0;
	for(int i=1;i<=n;i++)
	{
		LL tmp=k-d[i];
		if(tmp<0 || tmp%n) {printf("NO\n");return 0;}
		t+=tmp/n;
	}
	if(t!=k) printf("NO\n");
	else printf("YES\n");
	return 0;
}
```