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::{BufRead, Write, stdin, stdout};

fn solve<R: BufRead, W: Write>(mut input: R, mut output: W) {
    let mut s = String::new();
    input.read_line(&mut s).unwrap();
    let n: u32 = s.split_whitespace().next().unwrap().parse().unwrap();
    s.clear();
    let mut a: Vec<i64> = Vec::with_capacity(n as usize);
    input.read_line(&mut s).unwrap();
    for buf in s.split_whitespace() {
        a.push(buf.parse().unwrap());
    }
    let cnt: i64 = a.iter().fold(0, |acc, &x| acc + x) / ((1 + n) * n / 2) as i64;
    let mut result = (a.iter().fold(0, |acc, &x| acc + x) % ((1 + n) * n / 2) as i64) == 0;
    let first = a[0];
    for i in 1..a.len() {
        if (a[i] - i as i64 * cnt - first) % n as i64 != 0 {
            result = false;
            break;
        }
    }
    for i in a {
        if i < cnt || i > cnt * n as i64 {
            result = false;
            break;
        }
    }
    writeln!(output, "{}", if result { "YES" } else { "NO" }).unwrap();
}

fn main() {
    let stdin = stdin();
    solve(stdin.lock(), stdout());
}

#[cfg(test)]
mod tests {
    fn test(input: &str, output: &str) {
        let mut v = Vec::<u8>::new();
        ::solve(input.as_bytes(), &mut v);
        assert_eq!(String::from_utf8(v).unwrap(), output);
    }
    #[test]
    fn case1() {
        test("5
4 5 1 2 3\n",
             "YES\n");
    }
    #[test]
    fn case2() {
        test("5
6 9 12 10 8\n",
             "YES\n");
    }
    #[test]
    fn case3() {
        test("4
1 2 3 1\n",
             "NO\n");
    }
}