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 repl(i,a,b) for(int i=(int)(a);i<(int)(b);i++)
#define rep(i,n) repl(i,0,n)
#define replrev(i,a,b) for(int i=(int)(b)-1;i>=(int)(a);i--)
#define reprev(i,n) replrev(i,0,n)
#define repi(itr,ds) for(auto itr = ds.begin(); itr != ds.end(); ++itr)
#define mp make_pair
#define INF 2000000000
#define MOD 1000000007
#define PI 3.1415926536
typedef long long ll;
typedef pair<int, int> P;

int main() {
	int N;
	cin >> N;
	vector<ll> a(N), diff(N);
	ll sum = 0;
	rep(i, N) {
		cin >> a[i];
		sum += a[i];
	}
	if (sum % ((ll)N*(N + 1) / 2) != 0) {
		cout << "NO" << endl;
		return 0;
	}
	ll K = sum / ((ll)N*(N + 1) / 2);
	rep(i, N) {
		diff[i] = a[(i + 1) % N] - a[i];
	}
	ll num = 0;
	rep(i, N) {
		if ((K - diff[i]) % N != 0) {
			cout << "NO" << endl;
			return 0;
		}
		num += (K - diff[i]) / N;
	}
	if (num != K) {
		cout << "NO" << endl;
	}
	else {
		cout << "YES" << endl;
	}

	return 0;
}