#include<stdio.h>
typedef long long LL;
LL a[100005],d[100005];

int main()
{
    int i;
    LL N,t,S=0,S0,n;
    scanf("%lld", &n);
    S0=n*(n+1)/2;
    for(i=1;i<=n;i++){
        scanf("%lld", &a[i]);
        S+=a[i];
    }
    if(S%S0){
        printf("NO");
        return 0;
    }
    N=S/S0;
    for(i=1;i<=n;i++){
        d[i]=a[i%n+1]-a[i];
        t=N-d[i];
        if(t%n||t<0){
            printf("NO");
            return 0;
        }
    }
    printf("YES");
    return 0;
}

### Prompt

Generate 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<stdio.h>
typedef long long LL;
LL a[100005],d[100005];

int main()
{
    int i;
    LL N,t,S=0,S0,n;
    scanf("%lld", &n);
    S0=n*(n+1)/2;
    for(i=1;i<=n;i++){
        scanf("%lld", &a[i]);
        S+=a[i];
    }
    if(S%S0){
        printf("NO");
        return 0;
    }
    N=S/S0;
    for(i=1;i<=n;i++){
        d[i]=a[i%n+1]-a[i];
        t=N-d[i];
        if(t%n||t<0){
            printf("NO");
            return 0;
        }
    }
    printf("YES");
    return 0;
}

```