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>
#define abs(a) a>0?a:-a
using namespace std;
int a[100005];
int b[100005];

int main(){
	int n,m=0;
	int i,j,k;
	bool miss=false;
	cin>>n;
	for(i=0;i<n;i++){
		cin>>a[i];
	}
	for(i=0;i<n;i++){
		if(i==n-1){
			b[n-1]=a[n-1]-a[0];
		}else{
			b[i]=a[i]-a[i+1];
		}
		if(b[i]<0){
			j=(-b[i])/n+1;
			b[i]+=j*n;
		}
	}
	k=b[0]%n;
	for(i=0;i<n;i++){
		if(b[i]%n!=k){
			miss=true;
			break;
		}
	}
	if(miss){
		cout<<"NO"<<endl;
	}else{
		cout<<"YES"<<endl;
	}
	return 0;
}