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;
#define ll long long
#define ii pair<ll,ll>
#define iii pair<ll,ii>
#define endl '\n'

int n,m;
int arr[105][105];

vector<int> al[105];

int r[10005]; //which left node is it connected to
bool visited[10005];
bool dfs(int i){
  if (visited[i]) return false;
  visited[i]=true;
  for (int x=0;x<al[i].size();x++){
    if (r[al[i][x]/m]==-1 || dfs(r[al[i][x]/m])){
      r[al[i][x]/m]=i;
      return true;
    }
  }
  return false;
}

int ans[105][105];

int main(){
	ios::sync_with_stdio(0);
    cin.tie(0);
	
	cin>>n>>m;
	
	for (int x=0;x<n;x++)
		for (int y=0;y<m;y++) cin>>arr[x][y],arr[x][y]--;
		
	for (int x=0;x<n;x++)
		for (int y=0;y<m;y++)
			al[x].push_back(arr[x][y]);
			
	for (int x=0;x<m;x++){
		memset(r,-1,sizeof(r));
		
		for (int y=0;y<n;y++) {
			memset(visited,false,sizeof(visited));
			dfs(y);
		}
		
		for (int y=0;y<n;y++){
			for (int z=0;z<al[r[y]].size();z++){
				if (al[r[y]][z]/m==y){
					ans[r[y]][x]=al[r[y]][z];
					al[r[y]].erase(al[r[y]].begin()+z);
					break;
				}
			}
		}
	}
	
	for (int x=0;x<n;x++){
		for (int y=0;y<m;y++) cout<<ans[x][y]+1<<" ";
		cout<<endl;
	}
	
	vector<int> temp;
	
	for (int x=0;x<m;x++){
		temp.clear();
		for (int y=0;y<n;y++) temp.push_back(ans[y][x]);
		
		sort(temp.begin(),temp.end());
		
		for (int y=0;y<n;y++) ans[y][x]=temp[y];
	}
	
	for (int x=0;x<n;x++){
		for (int y=0;y<m;y++) cout<<ans[x][y]+1<<" ";
		cout<<endl;
	}
}