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: Wrong Answer)

use std::io::*;
use std::str::FromStr;

fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

fn show(a: &Vec<Vec<usize>>) {
    for row in a.iter() {
        for val in row.iter() {
            print!("{} ", val + 1)
        }
        println!();
    }
}

fn transpose(a: &Vec<Vec<usize>>) -> Vec<Vec<usize>> {
    let mut b = vec![vec![0; a.len()]; a[0].len()];
    for y in 0..a.len() {
        for x in 0..a[0].len() {
            b[x][y] = a[y][x];
        }
    }
    return b;
}

fn main() {
    let h: usize = read();
    let w: usize = read();
    let a: Vec<Vec<usize>> = (0..h)
        .map(|_| (0..w).map(|_| read::<usize>() - 1).collect())
        .collect();

    // first step
    let mut b = vec![vec![!0; w]; h];
    {
        let mut usd = vec![vec![false; h]; w];

        for y in 0..h {
            let mut done = vec![false; w];

            for _ in 0..w {
                let mut cnt = vec![0; w];

                for x in 0..w {
                    if done[x] {
                        cnt[x] = 1000000000;
                        continue;
                    }
                    for tx in 0..w {
                        if b[y][tx] != !0 || usd[tx][a[y][x] / w] {
                            continue;
                        }
                        cnt[x] += 1;
                    }
                }

                let x = (0..w).min_by_key(|&x| cnt[x]).unwrap();

                for tx in 0..w {
                    if b[y][tx] != !0 || usd[tx][a[y][x] / w] {
                        continue;
                    }
                    b[y][tx] = a[y][x];
                    usd[tx][a[y][x] / w] = true;
                    break;
                }
                done[x] = true;
            }
        }
    }
    let b = b;
    show(&b);

    // second step
    let mut t: Vec<Vec<usize>> = transpose(&b);
    for row in t.iter_mut() {
        row.sort();
    }
    let c = transpose(&t);
    show(&c);
}