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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<cstdio>
using namespace std;
int main(){
    long n,a[100000],sum=0,d[100000],k;
    scanf("%ld",&n);
    for(long i=0;i<n;i++){
        scanf("%ld",&a[i]);
        sum+=a[i];
    }
    if(sum%(n*(n+1)/2)){
        printf("NO\n");
        return 0;
    }
    k=sum/(n*(n+1)/2);
    for(long i=0;i<n;i++){
        d[i]=a[i]-a[(i+1)%n]+k;
    }
    sum=0;
    for(long i=0;i<n;i++){
        if(d[i]%n){
            printf("NO\n");
            return 0;
        }
        else{
            sum+=d[i];
        }
    }
    if(sum==n*k){
        printf("YES\n");
    }
    else{
        printf("NO\n");
    }
}