#include <bits/stdc++.h>

using namespace std;  

const int maxn = 1e5 + 10;

long long n;
long long a[maxn];
int main() {
	long long sum = 0;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%lld", &a[i]);
		sum += a[i];
	}
	a[n] = a[0];
	if(sum % (n * (n + 1) / 2) != 0) {
		printf("NO\n");
		return 0;
	}
	long long cnt = sum / (n * (n + 1) / 2);
	for(int i = 0; i < n; i++) {
		long long tmp = cnt + a[i] - a[i + 1];
		if(tmp < 0 || tmp % n != 0) {
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
} 
### Prompt

Please provide a cpp coded solution to the problem described below:

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;  

const int maxn = 1e5 + 10;

long long n;
long long a[maxn];
int main() {
	long long sum = 0;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%lld", &a[i]);
		sum += a[i];
	}
	a[n] = a[0];
	if(sum % (n * (n + 1) / 2) != 0) {
		printf("NO\n");
		return 0;
	}
	long long cnt = sum / (n * (n + 1) / 2);
	for(int i = 0; i < n; i++) {
		long long tmp = cnt + a[i] - a[i + 1];
		if(tmp < 0 || tmp % n != 0) {
			printf("NO\n");
			return 0;
		}
	}
	printf("YES\n");
	return 0;
} 
```