#include <bits/stdc++.h>
using namespace std;
int main() {
	int n;
	long long a[100005], sum=0, d, diff, x;
	bool flag = 1;
	scanf("%d", &n);
	if(n%2 == 0) d = n/2LL * (n+1LL);
	else d = (n+1LL)/2LL * n;
	for(int i=0; i<n; i++) {
		scanf("%lld", &a[i]);
		sum += a[i];
	}
	if(sum%d != 0) printf("NO\n");
	else {
		x = sum/d;
		for(int i=0; i<n; i++) {
			diff = a[(i+1)%n] - a[i];
			if(diff > x || (x-diff)%n > 0) {
				flag = 0;
				break;
			}
		}
		if(!flag) printf("NO\n");
		else printf("YES\n");
	}
	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 <bits/stdc++.h>
using namespace std;
int main() {
	int n;
	long long a[100005], sum=0, d, diff, x;
	bool flag = 1;
	scanf("%d", &n);
	if(n%2 == 0) d = n/2LL * (n+1LL);
	else d = (n+1LL)/2LL * n;
	for(int i=0; i<n; i++) {
		scanf("%lld", &a[i]);
		sum += a[i];
	}
	if(sum%d != 0) printf("NO\n");
	else {
		x = sum/d;
		for(int i=0; i<n; i++) {
			diff = a[(i+1)%n] - a[i];
			if(diff > x || (x-diff)%n > 0) {
				flag = 0;
				break;
			}
		}
		if(!flag) printf("NO\n");
		else printf("YES\n");
	}
	return 0;
}
```