#include <cstdio>
long long n, n2, s, k, d, a[100005];
int main() {
	int i;
	scanf("%d", &n);
	for (i=0; i<n; i++) scanf("%d", &a[i]), s += a[i];
	n2 = n * (n+1) / 2;
	if (s%n2) {puts("NO"); return 0;}
	k = s / n2;
	for (i=s=0; i<n; i++) {
		d = a[(i+1)%n] - a[i];
		if (k<d || (k-d)%n) {puts("NO"); return 0;}
		s += (k-d) / n;
	}
	puts(s==k ? "YES":"NO");
	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 <cstdio>
long long n, n2, s, k, d, a[100005];
int main() {
	int i;
	scanf("%d", &n);
	for (i=0; i<n; i++) scanf("%d", &a[i]), s += a[i];
	n2 = n * (n+1) / 2;
	if (s%n2) {puts("NO"); return 0;}
	k = s / n2;
	for (i=s=0; i<n; i++) {
		d = a[(i+1)%n] - a[i];
		if (k<d || (k-d)%n) {puts("NO"); return 0;}
		s += (k-d) / n;
	}
	puts(s==k ? "YES":"NO");
	return 0;
}
```