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 int n,a,b,i,m,min,max,dif,r=0;

  scanf("%lld",&n);
  if(n<2){puts("YES");return 0;};

  scanf("%lld %lld",&a,&b);
  if(a<b)a+=n;
  dif=a-b;
  //  printf("%lld %lld %lld\n",a,b,(a-b)%n);                                                    
  //  if(dif<0)dif+=n;                                                                           
  dif%=n;
  for(i=0;i<n-2;i++){
    a=b;
    scanf("%lld",&b);
    if(a<b)a+=n;

    if((a-b)%n-dif){puts("NO");return 0;};
    //    if(dif-(a-b)%n&&dif-(b-a)%b){puts("NO");return 0;};                                    
  }
  puts("YES");
}