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<iostream>
#include<vector>

using namespace std;

typedef long long ll;
typedef vector<ll> vl;

int main() {
	int n;
	cin >> n;
	vl A(n);
	for (int i = 0; i < n; i++) {
		cin >> A[i];
	}
	ll sum = 0;
	for (int i = 0; i < n; i++) {
		sum += A[i];
	}
	ll onesum = n*(n + 1) / 2;
	ll k = sum / onesum;
	for (int i = 0; i < n; i++) {
		ll d = A[(i + 1) % n] - A[i];
		d -= k;
		if (d > 0 || d%n != 0) {
			cout << "NO" << endl;
			return 0;
		}
	}
	cout << "Yes" << endl;
}