#include <bits/stdc++.h>

using namespace std;

typedef unsigned long long ull;
typedef long long ll;

int main(){
	ll n;
	cin >> n;
	ll a[n], sum=0;
	for(int i=0; i<n; i++){
		cin >> a[i];
		sum+=a[i];
	}
	
	if(sum%(n*(n+1)/2)!=0){
		cout << "NO" << endl;
		return 0;
	}

	ll k=sum/(n*(n+1)/2), d[n];
	for(int i=0; i<n-1; i++) d[i]=a[i]-a[i+1];
	d[n-1]=a[n-1]-a[0];

	ll cnt=0;
	for(int i=0; i<n; i++){
		if((d[i]+k)%n!=0 || d[i]+k<0){
			cout << "NO" << endl;
			return 0;
		}else cnt+=(d[i]+k)/n;
	}
	if(cnt==k) cout << "YES" << endl;
	else cout << "NO" << endl;
	return 0;
}
### Prompt

In cpp, your task is to solve 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;

typedef unsigned long long ull;
typedef long long ll;

int main(){
	ll n;
	cin >> n;
	ll a[n], sum=0;
	for(int i=0; i<n; i++){
		cin >> a[i];
		sum+=a[i];
	}
	
	if(sum%(n*(n+1)/2)!=0){
		cout << "NO" << endl;
		return 0;
	}

	ll k=sum/(n*(n+1)/2), d[n];
	for(int i=0; i<n-1; i++) d[i]=a[i]-a[i+1];
	d[n-1]=a[n-1]-a[0];

	ll cnt=0;
	for(int i=0; i<n; i++){
		if((d[i]+k)%n!=0 || d[i]+k<0){
			cout << "NO" << endl;
			return 0;
		}else cnt+=(d[i]+k)/n;
	}
	if(cnt==k) cout << "YES" << endl;
	else cout << "NO" << endl;
	return 0;
}
```