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)

// In the name of God

#include <bits/stdc++.h>
using namespace std;
const int N = 105;
int a[N][N], n, m, ans1[N][N], ans2[N][N];
bool change[N][N];
map<pair<int,int>,set<int>> mp;
int main() {
	ios_base::sync_with_stdio(0), cin.tie(0);
	cin >> n >> m;
	for(int i=0;i<n;i++) {
		for(int j=0;j<m;j++) {
			cin >> a[i][j];
		}
	}
	for(int i=0;i<n;i++) {
		for(int j=0;j<m;j++) {
			int go = a[i][j]/m;
			if(a[i][j]%m==0)go--;
			if(go!=i) mp[make_pair(go,i)].insert(a[i][j]);
		}
	}
	for(auto &k: mp) {
		pair<int,int> key = k.first;
		set<int> &v = k.second;
		if(key.first>key.second) continue;
		for(auto &val : v) {
			int ch = *mp[make_pair(key.second, key.first)].begin();
			int good=-1;
			for(int i=0;i<m;i++) if(change[key.first][i]==false&&change[key.second][i]==false) good=i;
			change[key.first][good]=true;
			change[key.second][good]=true;
			ans1[key.second][good]=val;
			ans1[key.first][good]=ch;
			mp[make_pair(key.second, key.first)].erase(ch);
		}
		v.clear();
	}
	for(int i=0;i<n;i++) {
		set<int> st;
		for(int j=0;j<m;j++) st.insert(a[i][j]);
		for(int j=0;j<m;j++) if(ans1[i][j]) st.erase(ans1[i][j]);
		for(int j=0;j<m;j++) if(ans1[i][j]==0) {
			ans1[i][j]=*st.begin();
			st.erase(*st.begin());
		}
	}
	for(int i=0;i<n;i++) {
		for(int j=0;j<m;j++) {
			cout << ans1[i][j] << " ";
		}
		cout << "\n";
	}
	for(int j=0;j<m;j++) {
		set<int> st;
		for(int i=0;i<n;i++) {
			st.insert(ans1[i][j]);
		}
		for(int i=0;i<n;i++) {
			ans2[i][j]=*st.begin();
			st.erase(*st.begin());
		}
	}
	for(int i=0;i<n;i++) {
		for(int j=0;j<m;j++) {
			cout << ans2[i][j] << " ";
		}
		cout << "\n";
	}
	return 0;
}