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 <cstdio>
#define N 100005
using namespace std;

typedef long long ll;
ll n, s1, s2, t, s, k, num, cnt, h, a[N], ans[N];
int main()
{
	ll i, j;
//	freopen ("input.txt", "r", stdin);
	cin >> n;
	if (n == 1) {
		cout << "YES" << endl;
		return 0;
	}
	for (i = 0; i < n; i++) {
		scanf ("%lld", &a[i]);
		s += a[i];
	}
	t = n * (n + 1) / 2;
	if (s % t > 0) {
		cout << "NO" << endl;
		return 0;
	}
	k = s / t;
	num = k - (a[0] - a[n - 1]);
	if (num % n > 0) {
		cout << "NO" << endl;
		return 0;
	}
	ans[0] = num / n;
	cnt = num / n;
	for (i = 0; i < n - 1; i++) {
		num = k - (a[i + 1] - a[i]);
		if (num % n > 0) {
			cout << "NO" << endl;
			return 0;
		}
		ans[i + 1] = num / n;
		cnt += num / n;
	}
	if (cnt != k) {
		cout << "NO" << endl;
		return 0;
	}
	j = 1;
	for (i = 0; i <n; i++) {
		h += ans[i] * j;
		if (j == 1) j = n;
		else j--;
	}
	if (h != a[0]) {
			cout << "NO" << endl;
			return 0;
	}
	for (i = 1; i < n; i++) {
		h += k;
		h -= ans[i] * n;
		if (h != a[i]) {
				cout << "NO" << endl;
				return 0;
		}
	}
	cout << "YES" << endl;
    return 0;
}