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 Rust (1.15.1) (judge verdict: Accepted)

use std::io;

fn main() {
  let mut input_line = String::new();
  io::stdin().read_line(&mut input_line).unwrap();
  let N = input_line.trim().parse::<usize>().unwrap();
  let Ni = N as isize;
  
  let mut input_line = String::new();
  io::stdin().read_line(&mut input_line).unwrap();
  let data = input_line.trim().split(" ").map(|x| x.parse::<isize>().unwrap()).collect::<Vec<_>>();

  let step = (N * (N+1) / 2) as isize;
  let total = data.iter().fold(0,|a, b| a + b);
  if total % step != 0 {
    println!("NO");
  }
  else {
    let K = total / step;
    let mut delta: Vec<isize> = Vec::with_capacity(N);
    for i in 0..N {
      delta.push(data[i] - data[(N + i-1) % N]);
    }
    let invalid = delta.iter().map(|x| (K-x < 0) || ((K-x) % Ni > 0)).fold(false,|a, b| a || b);
    if invalid {
      println!("NO");
    }
    else {
      let sum = delta.iter().map(|x| (K-x) / Ni).fold(0, |a, b| a + b);
      if sum == K {
        println!("YES");
      }
      else {
        println!("NO");
      }
    }
  }
}