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(void){
    int i,N,a,sa,sum=0;
    int A[100005]={0};
    scanf("%d",&N);
    for(i=0;i<N;i++){
        scanf("%d",&A[i]);
        sum+=A[i];
    }
    A[N]=A[0];
    if(sum%(N*(N+1)/2)!=0){printf("NO\n");return 0;}
    sa=sum/(N*(N+1)/2);
    for(i=0;i<N;i++){
        if((A[i]+sa-A[i+1])%N!=0){printf("NO\n");return 0;}
    }
    printf("YES\n");return 0;
}