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 <algorithm>
# include <vector>
# include <string>
# include <set>
# include <map>
# include <cmath>
# include <iomanip>
# include <functional>
# include <utility>
# include <stack>
# include <queue>
# include <list>
# include <tuple>
# include <unordered_map>
# include <numeric>
using namespace std;
using LL = long long;
using ULL = unsigned long long;
constexpr int INF = 2000000000;
constexpr int HINF = INF / 2;
constexpr double DINF = 100000000000000000.0;
constexpr long long LINF = 9223372036854775807;
constexpr long long HLINF = 4500000000000000000;
const double PI = acos(-1);
int dx[4] = { 0,1,0,-1 }, dy[4] = { 1,0,-1,0 };
# define ALL(x)    (x).begin(),(x).end()
# define UNIQ(c)   (c).erase(unique(ALL((c))), end((c)))
# define mp        make_pair
# define eb        emplace_back
# define REP       (i, a) for (int i = 0; i < (a); i++)
# define INIT      std::ios::sync_with_stdio(false);std::cin.tie(0);

LL n, a[101010], sum, sa;

int main() {
	cin >> n;
	for (int i = 0; i < n; i++) {
		cin >> a[i];
		sum += a[i];
		if (i) {
			sa += (a[i] - a[i - 1]);
		}
	}
	sa += a[0] - a[n - 1];
	if (sum % (((1 + n)*n) / 2) == 0 && sa == 0) {
		LL num = sum / (((1 + n)*n) / 2);
		for (int i = 1; i < n; i++) {
			if (a[i] - a[i - 1] >= 0 && a[i] - a[i - 1] != num) {
				cout << "NO" << endl;
				return 0;
			}
		}
		if (a[0] - a[n - 1] >= 0 && a[0] - a[n - 1] != num) {
			cout << "NO" << endl;
			return 0;
		}
		cout << "YES" << endl;
	}
	else {
		cout << "NO" << endl;
	}
}