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: Accepted)

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int N=100005;
int n;
ll s,t,d;
int a[N];
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++){scanf("%d",&a[i]);s+=a[i];}
	t=(n*(n+1ll))>>1;
	if(s%t!=0ll){printf("NO");return 0;}
	s/=t;t=0ll;a[n+1]=a[1];
	for(int i=1;i<=n;i++){
		d=s-(a[i+1]-a[i]);
		if(d%n!=0ll||d<0ll){printf("NO");return 0;}
		t+=d;
	}
	if(t==n*s)printf("YES");else printf("NO");
	return 0;
}