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: Wrong Answer)

use std::io::stdin;

#[allow(dead_code)]
fn input_ints() -> Vec<u64> {
    let mut st = String::new();
    match stdin().read_line(&mut st) {
        Ok(_) => st.split_whitespace().map(|s| s.parse().unwrap()).collect(),
        Err(_) => "".split_whitespace().map(|s| s.parse().unwrap()).collect(),
    }
}

#[allow(unused_variables)]
fn solve() {
    let n = input_ints()[0] as usize;
    let ns = input_ints();

    let total: u64 = ns.iter().fold(0, |acc, i| acc + i);
    let total_to_n: u64 = (n * (n + 1) / 2) as u64;

    let k = total / total_to_n;
    
    let mut d = vec![0i64; n];

    for i in 0..n-1 {
        d[i] = ns[i + 1] as i64 - ns[i] as i64 - k as i64;
    }
    d[n - 1] = ns[0] as i64 - ns[n - 1] as i64 - k as i64;

    if d.iter().all(|&x| x <= 0 && x.abs() % n as i64 == 0) {
        println!("NO");
    } else {
        println!("YES");
    }
}

fn main() {
    solve();
}