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: Time Limit Exceeded)

#include <bits/stdc++.h>

using namespace std;

typedef long long LL;

LL a[200000], cnt, b[200000], n, s, t;

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		s += a[i];
		b[i] = a[i];
		t += i + 1;
	}
	
	if (s % t != 0) {
		cout << "NO";
		return 0;
	}
	LL k = s / t;
	b[n] = b[0];
	
	if (n <= 1000) {
	
	for (int i = 0; i < n; i++) {
		if (k + (b[i] - b[i + 1]) > 0) {
			LL x = k + (b[i] - b[i + 1]);
			x /= n;
			for (int j = 0; j < n; j++) {
				a[(i + j + 1) % n] -= x * (j + 1); 
			}
		}
	}
	} else {
	
	for (int q = 1; q <= k; q++) {
		int p = 0;
		for (int i = 0; i < n; i++) {
			if (a[i] > a[(i + 1) % n]) {
				p = i;
				break;
			}		
		}
		for (int i = 0; i < n; i++) {
			a[(i + p + 1) % n] -= i + 1;
		}
	}
	}
	
	for (int i = 0; i < n; i++) {
		if (a[i] != 0) {
			cout << "NO";
			return 0;
		}
	}
	cout << "YES";
}