#include<cstdio>
int n, w[101000];
long long s, K;
int main(){
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&w[i]);
        s += w[i];
    }
    K = 1ll*n*(n+1)/2;
    if(s%K){
        puts("NO");
        return 0;
    }
    s/=K;
    long long t=0;
    for(i=1;i<=n;i++){
        if((w[i%n+1]%n-w[i]%n+n)%n != s%n){
            puts("NO");
            return 0;
        }
        if(w[i%n+1] - w[i] > s){
            puts("NO\n");
            return 0;
        }
        t += (s - (w[i%n+1] - w[i])) / n;
    }
    if(t!=s) puts("NO");
    else puts("YES");
}
### 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<cstdio>
int n, w[101000];
long long s, K;
int main(){
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&w[i]);
        s += w[i];
    }
    K = 1ll*n*(n+1)/2;
    if(s%K){
        puts("NO");
        return 0;
    }
    s/=K;
    long long t=0;
    for(i=1;i<=n;i++){
        if((w[i%n+1]%n-w[i]%n+n)%n != s%n){
            puts("NO");
            return 0;
        }
        if(w[i%n+1] - w[i] > s){
            puts("NO\n");
            return 0;
        }
        t += (s - (w[i%n+1] - w[i])) / n;
    }
    if(t!=s) puts("NO");
    else puts("YES");
}
```