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;

using ll = long long;

int main() {
    ll N;
    cin >> N;

    vector<ll> A(N);

    ll sum = 0;
    for (int i = 0; i < N; i++) {
        cin >> A[i];
        sum += A[i];
    }

    if ((2 * sum) % (N * (N + 1)) != 0) {
        cout << "NO" << endl;
        return 0;
    }

    ll k = (2 * sum) / (N * (N + 1)),
       life = k; // 操作回数

    for (int cur = 0; cur < N; cur++) {
        int next = (cur + 1) % N;

        ll diff = k - (A[next] - A[cur]);

        if (diff % N != 0) {
            cout << "NO" << endl;
            return 0;
        }

        life -= diff / N;
    }

    if (life == 0) cout << "YES" << endl;
    else cout << "NO" << endl;

    return 0;
}