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 <cstdio>
#include <algorithm>
const int N = 101;
int a[N][N];
int r[N * N]; // r[x] = row of element x
int p[N][N]; // p[x][y] = row of element y in column x

int next(int row)
{
	int col = 0;
	while (a[row][col] != -1)
		++col;
	return col;
}

int main()
{
	int n, m; scanf("%d%d", &n, &m);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j)
		{
			int x; scanf("%d", &x);
			r[x - 1] = i;
			a[i][j] = -1;
		}
	
	for (int i = 0; i < n * m; ++i)
	{
		int row = r[i];
		int col1 = i % m;
		int col2 = next(row);
		int val = i / m;
		if (col1 == col2)
		{
			a[row][col1] = val;
			p[col1][val] = row;
		} 
		else 
			while (val != -1)
			{
				p[col1][val] = row;
				std::swap(val, a[row][col1]);
				a[row][col2] = val;
				if (val != -1)
					std::swap(row, p[col2][val]);
			}
	}
	
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j)
		{
			int val = a[i][j] * m;
			for (int k = 0; k < m; ++k)
				if (r[val + k] == i)
				{
					a[i][j] = val + k;
					r[val + k] = -1;
					break;
				}
		}
	
	for (int i = 0; i < n; ++i, puts(""))
		for (int j = 0; j < m; ++j)
			printf("%d ", a[i][j] + 1);
	
	for (int i = 0; i < m; ++i)
		for (int j = 1; j < n; ++j)
			for (int k = 0; k < n - j; ++k)
				if (a[k][i] > a[k + 1][i])
					std::swap(a[k][i], a[k + 1][i]);
	
	for (int i = 0; i < n; ++i, puts(""))
		for (int j = 0; j < m; ++j)
			printf("%d ", a[i][j] + 1);
}