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 <bits/stdc++.h>
using namespace std;
using ll = long long;
using P = pair<int, int>;
const ll MOD = 1000000007;

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

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

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

    ll move = 0;
    for(int i=0;i<N;i++){
        if(d[i] < 0){
            ll new_move = (-d[i] + N - 2) / (N-1);
            d[i] += new_move * N;
            move += new_move;
        }
    }

    bool ok = true;
    for(int i=0;i<N;i++){
        if(d[i] - move != 0){
            ok = false;
            break;
        }
    }
    if(ok) cout << "Yes" << endl;
    else cout << "No" << endl;

    return 0;
}