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

long long int N, h, a[100001], s, t, i;

int main(){
    scanf("%lld%lld", &N, a);
    h = N * 1000000000LL;
    a[N] = *a;
    s = *a;
    for(; ++i < N;){
        scanf("%lld", a + i);
        s += a[i];
    }
    if(s % (N * (N + 1) / 2))return 0 & puts("NO");
    s /= (N * (N + 1) / 2);
    for(i = 0; i < N; ++i){
        t += (a[i] - a[i + 1] + s) / N;
        if((a[i] - a[i + 1] + s) < 0 || (a[i] - a[i + 1] + s) % N != 0)return 0 & puts("NO");
    }
    puts(((t + h) % N != (s + h) % N) ? "NO" : "YES");
    return 0;
}

### Prompt

Please formulate 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 <bits/stdc++.h>
using namespace std;

long long int N, h, a[100001], s, t, i;

int main(){
    scanf("%lld%lld", &N, a);
    h = N * 1000000000LL;
    a[N] = *a;
    s = *a;
    for(; ++i < N;){
        scanf("%lld", a + i);
        s += a[i];
    }
    if(s % (N * (N + 1) / 2))return 0 & puts("NO");
    s /= (N * (N + 1) / 2);
    for(i = 0; i < N; ++i){
        t += (a[i] - a[i + 1] + s) / N;
        if((a[i] - a[i + 1] + s) < 0 || (a[i] - a[i + 1] + s) % N != 0)return 0 & puts("NO");
    }
    puts(((t + h) % N != (s + h) % N) ? "NO" : "YES");
    return 0;
}

```