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 <bits/stdc++.h>

using namespace std;

const int MAXN = 1e2 + 5;

int n, m;
int a[MAXN][MAXN];

void read_input()
{
	scanf("%d %d", &n, &m);

	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			scanf("%d", &a[i][j]);
			a[i][j]--;
		}
		sort(a[i], a[i] + m);
	}
}
bool used[MAXN];
int par[MAXN];

int k;
vector <int> graph[MAXN];
int tr[MAXN];

bool try_kuhn(int p)
{
	if(used[p]) return false;
	used[p] = true;

	int z = graph[p].size();
	for(int i = 0; i < z; i++)
	{
		int to = graph[p][i];
		if(par[to] == -1 || try_kuhn(par[to]))
		{
			par[to] = p;
			tr[p] = to;
			return true;
		}
	}
	return false;
}

bool mark[MAXN][MAXN];
int b1[MAXN][MAXN], b2[MAXN][MAXN];

void solve()
{
	bool fl = true;
	for(int z = 0; z < m; z++)
	{
        memset(par, -1, sizeof(par));

		for(int i = 0; i < n; i++)
        {
            graph[i].clear();
            for(int j = 0; j < m; j++)
            {
                if(!mark[i][j])
                {
                    int temp = a[i][j] / m;
                    graph[i].push_back(temp);
                }
            }
        }

		for(int j = 0; j < n; j++)
		{
			memset(used, 0, sizeof(used));
			fl &= try_kuhn(j);
		}
		for(int i = 0; i < n; i++)
		{
			int temp = tr[i];

			for(int j = 0; j < m; j++)
            {
                if(a[i][j] / m == temp && !mark[i][j])
                {
                    mark[i][j] = true;
                    b1[i][z] = a[i][j];
                    break;
                }
            }
		}
	}

	if(!fl)
	{
	    printf("DSA");
		return;
	}
	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			printf("%d ", b1[i][j] + 1);
			b2[j][i] = b1[i][j] + 1;
		}
		printf("\n");
	}
	for(int i = 0; i < m; i++)
	{
		sort(b2[i], b2[i] + n);
	}

	for(int i = 0; i < n; i++)
	{
		for(int j = 0; j < m; j++)
		{
			printf("%d ", b2[j][i]);
		}
		printf("\n");
	}

}
int main()
{
	read_input();
	solve();

	return 0;
}