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)

//In the name of Allah 
#include <bits/stdc++.h>

using namespace std; 

#define F first
#define S second
#define MP make_pair
#define PB push_back
#define int long long

const int N=1e5+100;

int a[N];
int f(int x,int y,int mod){
	int k=x%mod-y%mod; 
	k+=2*mod; 
	return k%mod;
}

int32_t main(){
	ios::sync_with_stdio(0);cin.tie(0);	
	int n; 
	cin>>n; 
	int sum=0;
	for(int i=0;i<n;i++){
		cin>>a[i];
		sum+=a[i];
	}
	if(sum%(n*(n+1)/2LL)!=0)return cout<<"NO\n",0;
	if(n==1)return cout<<"YES\n",0;
	int k=f(a[0],a[1],n);
	for(int i=0;i<n-1;i++){
		if(f(a[i],a[i+1],n)!=k)return cout<<"NO\n",0;
	}
	if(f(a[n-1],a[0],n)!=k)return cout<<"NO\n",0;
	cout<<"YES\n";
	return 0;
}