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 C++14 (GCC 5.4.1) (judge verdict: Runtime Error)

#include<bits/stdc++.h>
typedef long long ll;
using namespace std;


int main(){
    int N,M;
    cin >> N >> M;
    int A[N][M];
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            cin >> A[i][j];
        }
        
    }
    int mp[N] = {0};
    int cnt = 0;
    int temp;
    for (int k = 0; k < M-1; k++)
    {
        for (int i = 0; i < N; i++)
            {
                mp[(A[i][k]-1)/M]++;
                if(mp[(A[i][k]-1)/M] > 1){
                    while(true){
                        cnt++;
                        if(mp[(A[i][k+cnt]-1)/M] == 0){
                            temp = A[i][k+cnt];
                            A[i][k+cnt] = A[i][k];
                            A[i][k] = temp;
                            break;
                        }
                    }
                    cnt = 0;
                }
            }
            for (int i = 0; i < N; i++)
            {
                mp[i] = 0;
            }
            
    }

    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            if(j != M-1) cout << A[i][j] << " ";
            else cout << A[i][j] << endl;
        }
        
    }
        
    int B[N][M];

    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            B[i][j] = (A[i][j]-1)/M;
        }
        
    }
    
    int C[N][M];
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            C[i][j] = A[B[i][j]][j];
        }
        
    }
    
    for (int i = 0; i < N; i++)
    {
        for (int j = 0; j < M; j++)
        {
            if(j != M-1) cout << C[i][j] << " ";
            else cout << C[i][j] << endl;
        }
        
    }
}