#include <bits/stdc++.h>
using namespace std;

long long n, a[100010], s, k;

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

### Prompt

Create a solution in Cpp for 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;

long long n, a[100010], s, k;

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

```