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;

long long n;
vector<long long> a;
template <class T>
void readVector(int n, vector<T> &v){
	for(int i=0; i<n; ++i){
		T t;
		cin >> t;
		v.push_back(t);
	}
}

int main(){
//	cin.tie(0);
//	ios::sync_with_stdio(false);
	cin >> n;
	readVector(n, a);
	long long s = 0;
	for(auto i : a)
		s += i;
	if(s % (n * (n+1) / 2) == 0){
		long long k = s * 2 / n / (n+1);
		bool f = true;
		for(int i=0; i<n; ++i){
			if((k - a[(i+1)%n] + a[i]) % n != 0)
				f = false;
			if(k - a[(i+1)%n] + a[i] < 0)
				f = false;
		}
//		cout << k << " " << tmp << "\n";
		cout << (f ? "YES\n" : "NO\n");
	}else{
		cout << "NO\n";
	}
}