Problem:
Score : 800 points

Problem Statement

Niwango has N cards, numbered 1,2,\ldots,N.
He will now arrange these cards in a row.

Niwango wants to know if there is a way to arrange the cards while satisfying all the N conditions below.
To help him, determine whether such a way exists. If the answer is yes, also find the lexicographically smallest such arrangement.

To the immediate right of Card 1 (if any) is NOT Card a_1.

To the immediate right of Card 2 (if any) is NOT Card a_2.

\vdots

To the immediate right of Card N (if any) is NOT Card a_N.

Constraints

2 \leq N \leq 10^{5}

1 \leq a_i \leq N

a_i \neq i

Input

Input is given from Standard Input in the following format:

N
a_1 a_2 \ldots a_N

Output

If no arrangements satisfy the conditions, print -1. If such arrangements exist, print the lexicographically smallest such arrangement, in the following format:

b_1 b_2 \ldots b_N

Here, b_i represents the i-th card from the left.

Sample Input 1

4
2 3 4 1

Sample Output 1

1 3 2 4

The arrangement (1,2,3,4) is lexicographically smaller than (1,3,2,4), but is invalid, since it violates the condition "to the immediate right of Card 1 is not Card 2."

Sample Input 2

2
2 1

Sample Output 2

-1

If no arrangements satisfy the conditions, print -1.

Sample Input 3

13
2 3 4 5 6 7 8 9 10 11 12 13 12

Sample Output 3

1 3 2 4 6 5 7 9 8 10 12 11 13

Solution in Rust (1.15.1) (judge verdict: Wrong Answer)

//https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8 より
macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
    ($($r:tt)*) => {
        let s = {
            use std::io::Read;
            let mut s = String::new();
            std::io::stdin().read_to_string(&mut s).unwrap();
            s
        };
        let mut iter = s.split_whitespace();
        input_inner!{iter, $($r)*}
    };
}

macro_rules! input_inner {
    ($iter:expr) => {};
    ($iter:expr, ) => {};
    ($iter:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($iter, $t);
        input_inner!{$iter $($r)*}
    };
}

macro_rules! read_value {
    ($iter:expr, ( $($t:tt),* )) => {
        ( $(read_value!($iter, $t)),* )
    };
    ($iter:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($iter, $t)).collect::<Vec<_>>()
    };
    ($iter:expr, chars) => {
        read_value!($iter, String).chars().collect::<Vec<char>>()
    };
    ($iter:expr, usize1) => {
        read_value!($iter, usize) - 1
    };
    ($iter:expr, $t:ty) => {
        $iter.next().unwrap().parse::<$t>().expect("Parse error")
    };
}

//

fn valid(a: &[usize], ans: &[usize]) -> bool {
    for i in 1..ans.len() {
        if a[ans[i - 1]] == ans[i] {
            return false;
        }
    }
    true
}

fn dfs(ans: &mut [usize], a: &[usize], k: usize) -> bool {
    if k == 0 {
        let res = valid(a, ans);
        if res {
            let mut out = String::new();
            for a in ans[1..].iter() {
                out.push_str(&format!("{} ", *a));
            }
            out.pop();
            println!("{}", out);
        }
        return res;
    }
    if dfs(ans, a, k - 1) {
        return true;
    }
    let n = a.len() - 1;
    for i in (n - k + 1)..(n + 1) {
        ans.swap(n - k, i);
        if dfs(ans, a, k - 1) {
            return true;
        }
    }
    let t = ans[n - k];
    for i in (n - k + 1)..(n + 1) {
        ans[i - 1] = ans[i];
    }
    ans[n] = t;
    false
}

fn run() {
    input! {
        n: usize,
        a: [usize; n],
    }
    if n == 2 {
        println!("-1");
        return;
    }
    let mut a = a;
    a.reverse();
    a.push(0);
    a.reverse();
    if n == 3 {
        for &(x, y, z) in [(1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1)].iter() {
            if valid(&a, &[x, y, z]) {
                println!("{} {} {}", x, y, z);
                return;
            }
        }
        println!("-1");
        return;
    }
    let mut ans = vec![0; n + 1];
    let mut set = std::collections::BTreeSet::new();
    for i in 1..(n + 1) {
        set.insert(i);
    }
    for i in 1..(n + 1 - 4) {
        let mut v = 0;
        for &k in set.iter() {
            if a[ans[i - 1]] != k {
                v = k;
                break;
            }
        }
        set.remove(&v);
        ans[i] = v;
    }
    let b: Vec<usize> = set.into_iter().collect();
    ans[n - 3] = b[0];
    ans[n - 2] = b[1];
    ans[n - 1] = b[2];
    ans[n - 0] = b[3];
    dfs(&mut ans, &a, 3);
}

fn main() {
    run();
}