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: Accepted)

#include <bits/stdc++.h>

using namespace std;

using int64 = long long;
const int64 INF = 1LL << 62;

int main()
{
  int64 N, A[100000];

  cin >> N;
  int64 all = 0;
  for(int i = 0; i < N; i++) {
    cin >> A[i];
    all += A[i];
  }

  int64 add = N * (N + 1) / 2;
  if(all % add != 0) {
    cout << "NO" << endl;
    return (0);
  }
  all /= add;

  vector< int64 > sa(N);
  for(int i = 0; i < N; i++) {
    sa[i] = A[(i + 1) % N] - A[i];
  }

  // 全体を -1 してどこかの要素に +N する操作を all 回できる
  // どうせ -1 が all 回されるので最初に -all しておく

  for(int i = 0; i < N; i++) {
    sa[i] -= all;
  }

  // あとは自明なので
  for(int i = 0; i < N; i++) {
    if(sa[i] > 0 || sa[i] % N != 0) {
      cout << "NO" << endl;
      return (0);
    }
  }

  cout << "YES" << endl;
}