#include <cstdio>

int p[100010];

int main(){
  int N; scanf("%d", &N);
  for(int i = 0; i < N; i++) scanf("%d", &p[i]);

  long long S = 0;
  for(int i = 0; i < N; i++) S += p[i];

  if(S % (1LL * N * (N + 1) / 2) != 0){
    puts("NO"); return 0;
  }

  long long T = S / (1LL * N * (N + 1) / 2);

  bool v = true;

  p[N] = p[0];
  for(int i = 0; i < N; i++){
    int D = p[i + 1] - p[i];
    long long V = T - D;
    if(V < 0 || V % N != 0) v = false;
  }

  puts(v ? "YES" : "NO");
  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 <cstdio>

int p[100010];

int main(){
  int N; scanf("%d", &N);
  for(int i = 0; i < N; i++) scanf("%d", &p[i]);

  long long S = 0;
  for(int i = 0; i < N; i++) S += p[i];

  if(S % (1LL * N * (N + 1) / 2) != 0){
    puts("NO"); return 0;
  }

  long long T = S / (1LL * N * (N + 1) / 2);

  bool v = true;

  p[N] = p[0];
  for(int i = 0; i < N; i++){
    int D = p[i + 1] - p[i];
    long long V = T - D;
    if(V < 0 || V % N != 0) v = false;
  }

  puts(v ? "YES" : "NO");
  return 0;
}

```