Problem:
Score : 1100 points

Problem Statement

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:

First, for each of the N rows, rearrange the numbers written in it as you like.

Second, for each of the M columns, rearrange the numbers written in it as you like.

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.

Sample Input 1

3 2
2 6
4 3
1 5

Sample Output 1

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

Sample Input 2

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

Sample Output 2

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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define st first
#define nd second

using namespace std;

void debug_out() { cerr << endl; }
template<class T> ostream& prnt(ostream& out, T v) { out << v.size() << '\n'; for(auto e : v) out << e << ' '; return out;}
template<class T> ostream& operator<<(ostream& out, vector <T> v) { return prnt(out, v); }
template<class T> ostream& operator<<(ostream& out, set <T> v) { return prnt(out, v); }
template<class T1, class T2> ostream& operator<<(ostream& out, map <T1, T2> v) { return prnt(out, v); }
template<class T1, class T2> ostream& operator<<(ostream& out, pair<T1, T2> p) { return out << '(' << p.st << ' ' << p.nd << ')'; }
template <typename Head, typename... Tail> void debug_out(Head H, Tail... T) { cerr << " " << H; debug_out(T...);}

#define dbg(...) cerr << #__VA_ARGS__ << " ->", debug_out(__VA_ARGS__)
#define dbg_v(x, n) do{cerr<<#x"[]: ";for(int _=0;_<n;++_)cerr<<x[_]<<" ";cerr<<'\n';}while(0)
#define dbg_ok cerr<<"OK!\n"

const int N = 101;
int n, m, a[N][N];

struct BipartiteMatcher {
	vector<vector<int>> G;
	vector<int> L, R, Viz;
 
	BipartiteMatcher(int n, int m) {
		G.resize(n);
		L.resize(n, -1);
		R.resize(m, -1);
		Viz.resize(n);
	}
 
	void AddEdge(int a, int b) {
		G[a].push_back(b);
	}
 
	bool Match(int node) {
		if(Viz[node]) 
			return false;
		Viz[node] = true;
	
		for(auto vec : G[node]) {
			if(R[vec] == -1 || Match(R[vec])) {
				L[node] = vec;
				R[vec] = node;
				return true;
			}
		}
 
		return false;
	}
	void Solve() {
		bool ok = true;
		while(ok) {
			ok = false;
			fill(Viz.begin(), Viz.end(), 0);
			for(int i = 0; i < L.size(); ++i)
				if(L[i] == -1)
					ok |= Match(i);
		}
	}
};
 

int main() {
	ios_base::sync_with_stdio(false);

	cin >> n >> m;
	for(int i= 0; i < n; i++)
		for(int j = 0; j < m; j++) {
			cin >> a[i][j];
			a[i][j]--;
		}

	for(int j = 0; j < m; j++) {
		BipartiteMatcher bm(n, n);
		dbg(j);
		for(int i = 0; i < n; i++)
			for(int k = j; k < m; k++)
				bm.AddEdge(i, a[i][k] / m);
		// dbg_ok;
		bm.Solve();
		// dbg_ok;
		for(int i = 0; i < n; i++) {
			int sch = bm.L[i];
			dbg(i, sch);
			for(int k = j; k < m; k++)
				if(a[i][k] / m == sch) {
					swap(a[i][k], a[i][j]);
					break;
				}
		}
	}

	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			cout << a[i][j] << " \n"[j == m - 1];


	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++) {
			swap(a[i][j], a[a[i][j] / m][j]);
		}

	for(int i = 0; i < n; i++)
		for(int j = 0; j < m; j++)
			cout << a[i][j]  << " \n"[j == m - 1];
}