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 Ruby (2.3.3) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
using namespace std;
 
#define int long long
#define REP(i,n) for(int i=0;i<n;i++)
int n,a[100002],d[100002],sum=0,dsum=0;
 
int mod(int a,int b){return (a%b+b)%b;}
 
signed main(){
    cin>>n;
    int N=n*(n+1)/2;
    REP(i,n)cin>>a[i],sum+=a[i];
    if(sum%N){
        cout<<"NO"<<endl;
        return 0;
    }
    REP(i,n)d[i]=a[(i+1)%n]-a[i],dsum+=d[i];
    if(dsum){
        cout<<"NO"<<endl;
        return 0;
    }
    int t=sum/N;
    REP(i,n){
        if(d[i]<t&&mod(d[i],n)!=t){
            cout<<"NO"<<endl;
            return 0;
        }
    }
    cout<<"YES"<<endl;
    return 0;
}