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>
using namespace std;

int N;
long A[100000];

void Diff() {
  long tmp = A[0];
  for(int i = 1; i < N; ++i) {
    A[i - 1] = A[i] - A[i - 1];
  }
  A[N - 1] = tmp - A[N - 1];
}

int main() {
  cin >> N;
  long sum = 0;
  for(int i = 0; i < N; ++i) {
    cin >> A[i];
    sum += A[i];
  }
  Diff();
  Diff();
  bool ok = true;
  long t = 0;
  for(int i = 0; i < N; ++i) {
    if(A[i] % N != 0) ok = false;
    t += A[i] / N * i;
  }
  if((sum - t * N * (N + 1) / 2) % N != 0) ok = false;
  
  cout << (ok ? "YES" : "NO") << endl;
}