Problem:
Score : 500 points

Problem Statement

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.

Sample Input 1

5
4 5 1 2 3

Sample Output 1

YES

All the stones can be removed in one operation by selecting the second box.

Sample Input 2

5
6 9 12 10 8

Sample Output 2

YES

Sample Input 3

4
1 2 3 1

Sample Output 3

NO

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <stdio.h>

int main()
{
  int N;
  int tmp;
  int k;
  long A[100001];
  long d[100001];
  int i;
  long sum = 0;

  scanf("%d", &N);
  for(i = 0; i < N; i++) {
    scanf("%ld", &A[i]);
    sum += A[i];
  }
  tmp = N*(N+1)*0.5;
  if(sum%tmp == 0) {
    k = sum/tmp;
  } else {
    printf("NO\n");
    return 0;
  }
  
  for(i = 0; i < N-1; i++) {
    d[i] = A[i] - A[i+1] + k;
  }
  d[N-1] = A[N-1] - A[0] + k;

  for(i = 0; i < N; i++) {
    if(d[i]%N != 0) {
      printf("NO\n");
      return 0;
    }
  }

  printf("YES\n");
  return 0;
}