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<bits/stdc++.h>
 
using namespace std;
 
#define pb push_back
#define mp make_pair
#define F first
#define S second
 
#define fo(i, n) for(int i = 1; i <= n; ++i)
#define rep(i, a, b) for(int (i) = (a); (i) < (b); ++(i))
 
typedef long long ll;
typedef pair <int, int> pii;
typedef pair <ll, ll> pll;
 
const int N = 200200;
const int mod = 1e9 + 7;
const ll INF = 1e18;
ll a[N],b[N],pref[N];
ll cnt[N];
 
int main()
{
	ios::sync_with_stdio(0); cin.tie(0);
	int n;
	cin>>n;
	for(int i=1;i<=n;++i)
	{
		cin>>a[i];
		pref[i]=pref[i-1]+a[i];
	}
	ll k=(n*1ll*(n+1)/2);
	if(pref[n]%k!=0)
		return cout<<"NO",0;
	ll x=pref[n]/k;
	a[n+1]=a[1];
	ll cc=0;
	for(int i=1;i<=n;++i)
	{
		ll f=x+(a[i]-a[i+1]);
		cnt[i]=f/n;
		cc+=f/n;
		cnt[i+n]=cnt[i];
	}
	ll sum=0;
	ll d=0,done=0;
	ll del=0;
	for(int i=1;i<=n+n;++i)
	{
		int cur=(i>n?i-n:i);
		sum+=d*i-done;
		b[cur]=sum;
		if(i==n)
			del=sum;
		done+=i*cnt[i];
		if(i>n)done-=cur*cnt[cur];
		d+=cnt[i];
		if(i>n)	d-=cnt[cur];
	}
	for(int i=n;i>=1;--i)
		b[i]-=b[i-1];
	b[1]-=del;
	for(int i=1;i<=n;++i)
		if(b[i]!=a[i])
			return cout<<"NO",0;
	cout<<"YES";
	return 0;	
}