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>

const int MAXN = 110;
const int MAXE = MAXN * MAXN;
int A[MAXN][MAXN], at[MAXE];
std::queue<int> G[MAXN][MAXN];
int col[2][MAXN][MAXN], ans[MAXE];
void dfs(int l, int r, int x, int y, int cx, int cy) {
	col[l][x][cx] = y;
	if (int v = col[r][y][cx]) {
		col[l][v][cx] = 0;
		col[r][y][cx] = x;
		dfs(r, l, y, v, cy, cx);
	} else col[r][y][cx] = x;
}
void adde(int x, int y) {
	int ca, cb;
	static int bak = 0;
	G[x][y].push(++bak);
	for (ca = 1; col[0][x][ca]; ++ca) ;
	for (cb = 1; col[1][y][cb]; ++cb) ;
	dfs(0, 1, x, y, ca, cb);
}
int B[MAXN][MAXN];
int main() {
	std::ios_base::sync_with_stdio(false), std::cin.tie(0);
	int n, m; std::cin >> n >> m;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			std::cin >> A[i][j], at[A[i][j]] = i;
	for (int i = 1; i <= n * m; ++i)
		adde(at[i], (i - 1) / m + 1);
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			if (int v = col[0][i][j])
				ans[G[i][v].front()] = j, G[i][v].pop();
	for (int i = 1; i <= n * m; ++i)
		B[at[i]][ans[i]] = i;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			std::cout << B[i][j] << (" \n" [j == m]);
	for (int i = 1; i <= n * m; ++i)
		B[(i - 1) / m + 1][ans[i]] = i;
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			std::cout << B[i][j] << (" \n" [j == m]);
	return 0;
}