#include<iostream>
#include<vector>
#include<algorithm>
#include<numeric>

using namespace std;

int main(){
	long long n, ans = 0;
	cin >> n;
	vector<long long> a(n), d(n);
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	long long k = accumulate(begin(a), end(a), 0LL), m = n * (n + 1) / 2;
	if(k % m != 0){
		cout << "NO" << endl;
		return 0;
	}
	k /= m;
	bool f = true;
	for(int i = 0; i < n; i++){
		d[i] = a[(i + 1) % n] - a[i] - k;
		if(d[i] % n != 0 || d[i] > 0){
			f = false;
			break;
		}
	}
	cout << (f? "YES":"NO") << endl;
	return 0;
}
### Prompt

Your challenge is to write 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<iostream>
#include<vector>
#include<algorithm>
#include<numeric>

using namespace std;

int main(){
	long long n, ans = 0;
	cin >> n;
	vector<long long> a(n), d(n);
	for(int i = 0; i < n; i++){
		cin >> a[i];
	}
	long long k = accumulate(begin(a), end(a), 0LL), m = n * (n + 1) / 2;
	if(k % m != 0){
		cout << "NO" << endl;
		return 0;
	}
	k /= m;
	bool f = true;
	for(int i = 0; i < n; i++){
		d[i] = a[(i + 1) % n] - a[i] - k;
		if(d[i] % n != 0 || d[i] > 0){
			f = false;
			break;
		}
	}
	cout << (f? "YES":"NO") << endl;
	return 0;
}
```