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(){
  long long n,d[100010],s=0,m,i,p;
  scanf("%lld",&n);
  m=(n*(n+1))/2;
  for(i=0;i<n;i++){
    scanf("%lld",&d[i]);
    s+=d[i];
  }
  if(s%m)goto END;
  s/=m;//printf("%lld\n",s);
  p=d[0];
  for(i=1;i<n;i++)d[i-1]+=s-d[i];
  d[n-1]+=s-p;
  //for(i=0;i<n;i++)printf("%lld ",d[i]);printf("\n");
  for(i=0;i<n;i++){
    if(d[i]%n)goto END;
  }
  printf("YES\n");
  return 0;
 END:;
  printf("NO\n");
  return 0;
}