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], b[MAXN][MAXN];
vector <int> gp[MAXN][MAXN];
vector <int> adj[MAXN];
bool bio[MAXN];
int conn[MAXN];

void load() {
	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]--;
		}
}

bool dfs(int x) {
	bio[x] = true;
	for (auto it : adj[x])
		if (conn[it] == -1 || (!bio[conn[it]] && dfs(conn[it]))) {
			conn[it] = x;
			return true;
		}
	return false;
}

void matching(int col) {
	for (int i = 0; i < N; i++) {
		adj[i].clear();
		for (int j = 0; j < N; j++)
			if (!gp[i][j].empty())
				adj[i].push_back(j);
	}
	
	memset(conn, -1, sizeof conn);
	for (int i = 0; i < N; i++) {
		memset(bio, false, sizeof bio);
		dfs(i);
	}
	
	for (int i = 0; i < N; i++) {
		b[conn[i]][col] = gp[conn[i]][i].back();
		gp[conn[i]][i].pop_back();
	}
}

void output(int mat[MAXN][MAXN]) {
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < M; j++)
			printf("%d ", mat[i][j] + 1);
		puts("");
	}
}

void solve() {
	for (int i = 0; i < N; i++)
		for (int j = 0; j < M; j++)
			gp[i][a[i][j] / M].push_back(a[i][j]);
			
	for (int j = 0; j < M; j++)
		matching(j);
	
	output(b);
	
	for (int j = 0; j < M; j++) {
		vector <int> col;
		for (int i = 0; i < N; i++)
			col.push_back(b[i][j]);
		sort(col.begin(), col.end());
		for (int i = 0; i < N; i++)
			b[i][j] = col[i];
	}
	
	output(b);
}

int main() {
	load();
	solve();
	return 0;
}