#include <bits/stdc++.h>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;

const int N = 100010;
int n,a[N];
int main(){
	scanf("%d",&n);
	ll sum  =0;
	for(int i =0;i<n;i++) scanf("%d",a+i),sum+=a[i];
	ll t =0;
	
	for(int i =0;i<n;i++) t+=a[(i+1)%n]-a[i];
	if(t==0 && sum%(1ll*(1+n)*n/2)==0) {
		ll k = sum/(1ll*(1+n)*n/2);
		bool pd =true;
		for(int i =0;i<n;i++) if((k-(a[(i+1)%n]-a[i]))%n!=0 || k-(a[(i+1)%n]-a[i])<0) pd=  false;
		if(pd) puts("YES") ;else puts("NO");
	} else puts("NO");
	return 0;
}
### Prompt

Please formulate 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>
#define X first
#define Y second
#define mp make_pair
using namespace std;
typedef pair<int,int> PII;
typedef long long ll;

const int N = 100010;
int n,a[N];
int main(){
	scanf("%d",&n);
	ll sum  =0;
	for(int i =0;i<n;i++) scanf("%d",a+i),sum+=a[i];
	ll t =0;
	
	for(int i =0;i<n;i++) t+=a[(i+1)%n]-a[i];
	if(t==0 && sum%(1ll*(1+n)*n/2)==0) {
		ll k = sum/(1ll*(1+n)*n/2);
		bool pd =true;
		for(int i =0;i<n;i++) if((k-(a[(i+1)%n]-a[i]))%n!=0 || k-(a[(i+1)%n]-a[i])<0) pd=  false;
		if(pd) puts("YES") ;else puts("NO");
	} else puts("NO");
	return 0;
}
```