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>
#include <numeric>
using namespace std;

typedef long long ll;

int main(void) {

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

    int once = N*(N+1) / 2;
    int k = accumulate(A.begin(), A.end(), 0L) / once;

    vector<ll> d(N);
    if (N != 1) {
        for (int i = 0; i < N; i++) {
            d[i] = A[(i+1)%N] - A[i];
        }
    } else {
        d[0] = -A[0];
    }

    for (int i = 0; i < N; i++) {
        while (d[i] < 0) {
            d[i] += N;
            k--;
        }
    }

    bool same = true;
    for (int i = 1; i < N; i++) {
        if (d[i] != d[i-1]) same = false;
    }

    if (k == 0 && same) {
        cout << "YES" << endl;
    } else {
        cout << "NO" << endl;
    }

    return 0;
}