Problem:
Score : 1100 points

Problem Statement

We have a grid with N rows and M columns of squares.
Each integer from 1 to NM is written in this grid once.
The number written in the square at the i-th row from the top and the j-th column from the left is A_{ij}.

You need to rearrange these numbers as follows:

First, for each of the N rows, rearrange the numbers written in it as you like.

Second, for each of the M columns, rearrange the numbers written in it as you like.

Finally, for each of the N rows, rearrange the numbers written in it as you like.

After rearranging the numbers, you want the number written in the square at the i-th row from the top and the j-th column from the left to be M\times (i-1)+j.
Construct one such way to rearrange the numbers. The constraints guarantee that it is always possible to achieve the objective.

Constraints

1 \leq N,M \leq 100

1 \leq A_{ij} \leq NM

A_{ij} are distinct.

Input

Input is given from Standard Input in the following format:

N M
A_{11} A_{12} ... A_{1M}
:
A_{N1} A_{N2} ... A_{NM}

Output

Print one way to rearrange the numbers in the following format:

B_{11} B_{12} ... B_{1M}
:
B_{N1} B_{N2} ... B_{NM}
C_{11} C_{12} ... C_{1M}
:
C_{N1} C_{N2} ... C_{NM}

Here B_{ij} is the number written in the square at the i-th row from the top and the j-th column from the left after Step 1, and C_{ij} is the number written in that square after Step 2.

Sample Input 1

3 2
2 6
4 3
1 5

Sample Output 1

2 6
4 3
5 1
2 1
4 3
5 6

Sample Input 2

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

Sample Output 2

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

Solution in Rust (1.15.1) (judge verdict: Accepted)

macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String {
            bytes
                .by_ref()
                .map(|r| r.unwrap() as char)
                .skip_while(|c| c.is_whitespace())
                .take_while(|c| !c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

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

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

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, [ $t:tt ]) => {
        {
            let len = read_value!($next, usize);
            (0..len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
        }
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

#[allow(unused_imports)]
use std::cmp::{min, max};
use std::collections::{HashSet, VecDeque};

fn main() {
    input!{
        n: usize,
        m: usize,
        a: [[usize; m]; n]
    }
    let mut es = vec![vec![vec![]; n]; n];
    for i in 0..n {
        for j in 0..m {
            es[i][(a[i][j] - 1) / m].push(a[i][j]);
        }
    }
    let mut b = vec![vec![0; m]; n];
    let mut c = vec![vec![0; m]; n];
    for i in 0..m {
        let mut g = vec![HashSet::new(); 2 * n + 2];
        for i in 0..n {
            for j in 0..n {
                if !es[i][j].is_empty() {
                    g[i].insert(n + j);
                }
            }
            g[2 * n].insert(i);
            g[n + i].insert(2 * n + 1);
        }
        let fpath = max_flow(2 * n, 2 * n + 1, &g);
        for &(u, v) in &fpath {
            if u < n {
                b[u][i] = es[u][v - n].pop().unwrap();
            }
        }
    }
    for i in 0..n {
        for j in 0..m {
            c[(b[i][j] - 1) / m][j] = b[i][j];
        }
    }
    print_tab(&b);
    print_tab(&c);
}

fn max_flow(s: usize, t: usize, g: &Vec<HashSet<usize>>) -> Vec<(usize, usize)> {
    let mut res = g.clone();
    let mut set = HashSet::new();
    while let Some(p) = bfs(s, t, &res) {
        for &(u, v) in &p {
            if set.contains(&(v, u)) {
                set.remove(&(v, u));
            } else {
                set.insert((u, v));
            }
            res[u].remove(&v);
            res[v].insert(u);
        }
    }
    set.iter().cloned().collect()
}

fn bfs(s: usize, t: usize, g: &Vec<HashSet<usize>>) -> Option<Vec<(usize, usize)>> {
    let n = g.len();
    let mut pre = vec![n; n];
    let mut q = VecDeque::new();
    q.push_back(s);
    while let Some(u) = q.pop_front() {
        if u == t {
            break;
        }
        for &v in g[u].iter() {
            if pre[v] == n {
                pre[v] = u;
                q.push_back(v);
            }
        }
    }
    if pre[t] == n {
        None
    } else {
        let mut ret = vec![];
        let mut p = t;
        while p != s {
            ret.push((pre[p], p));
            p = pre[p];
        }
        ret.reverse();
        Some(ret)
    }
}

fn print_tab(a: &Vec<Vec<usize>>) {
    for i in 0..a.len() {
        print!("{}", a[i][0]);
        for j in 1..a[0].len() {
            print!(" {}", a[i][j]);
        }
        println!("");
    }
}