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<iostream>
#include<vector>
#include<string>
using namespace std;

#define FOR(k,m,n) for(int (k)=(m);(k)<(n);(k)++)
#define rep(i,n) FOR((i),0,(n))

typedef long long ll;

const int INF=1e9+7;
const int MAX_N=1e5+5;

ll A[MAX_N];
int main(){
	ll n;
	cin>>n;
	rep(i,n)cin>>A[i];
	ll sum=0;
	rep(i,n)sum+=A[i];
	if(((2*sum)%(n*(n+1)))!=0){
		cout<<"NO"<<endl;
		return 0;
	}
	ll k=2*sum/(n*(n+1));
	ll a=(A[n-1]-A[0]+k)/n;
	rep(i,n-1){
		if((A[i]-A[i+1]+k)%n==0)a+=((A[i]-A[i+1]+k)/n);
		else{
			cout<<"NO"<<endl;
			return 0;
		}
	}
cout<<"NO"<<endl;
}