Problem: 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:

1. First, for each of the N rows, rearrange the numbers written in it as you like.
2. Second, for each of the M columns, rearrange the numbers written in it as you like.
3. 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.

Examples

Input

3 2
2 6
4 3
1 5


Output

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


Input

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


Output

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 CPP):

Solution:
#include <iostream>
#include <vector>
#include <set>
#include <map>
#include <algorithm>


using namespace std;

int n, m;
pair<int, int> a[100][100];
int b[100][100];

vector<int> bg[200];
int used[200];
int pairing[200];

bool kuhn(int v)
{
    if (used[v]) return 0;
    used[v] = 1;
    for (int i = 0; i < bg[v].size(); i++)
    {
        int u = bg[v][i];
        if (pairing[u] == -1 || kuhn(pairing[u]))
        {
            pairing[u] = v;
            return 1;
        }
    }
    return 0;
}

int main()
{
    cin >> n >> m;
    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            cin >> a[i][j].first;
            a[i][j].second = (a[i][j].first - 1) / m;
        }
    }

    for (int j = 0; j < m; j++)
    {
        for (int i = 0; i < 200; i++) bg[i].resize(0);

        for (int num = 0; num < n; num++)
        {
            for (int row = 0; row < n; row++)
            {
                int t = 0;
                for (int j2 = j; j2 < m; j2++)
                {
                    if (a[row][j2].second == num)
                    {
                        t = 1;
                    }
                }
                if (t)
                {
                    bg[num].push_back(100 + row);
                    bg[100 + row].push_back(num);
                }
            }
        }

        for (int i = 100; i < 200; i++) pairing[i] = -1;

        for (int num = 0; num < n; num++)
        {
            fill(used, used + n, 0);
            kuhn(num);
        }

        for (int row = 0; row < n; row++)
        {
            //cout << "number " << pairing[row + 100] << " from row " << row << endl;
            int x = pairing[row + 100];
            for (int j2 = j; j2 < m; j2++)
            {
                if (a[row][j2].second == x)
                {
                    swap(a[row][j2], a[row][j]);
                    break;
                }
            }
        }
    }

    for (int i = 0; i < n; i++)
    {
        for (int j = 0; j < m; j++)
        {
            cout << a[i][j].first << " ";
            b[j][i] = a[i][j].first;
        }
        cout << "\n";
    }

    for (int i = 0; i < m; i++)
        sort(b[i], b[i] + n);

    for (int j = 0; j < n; j++)
    {
        for (int i = 0; i < m; i++)
        {
            cout << b[i][j] << " ";
        }
        cout << "\n";
    }
}