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 <iostream>
#include <algorithm>
#include <string>
#include <vector>
#include <cmath>
#include <map>
#include <queue>
#include <iomanip>
#include <set>
#define MOD 1000000007
#define mkp make_pair
typedef long long ll;
using namespace std;

struct BipartiteMatching {
  vector<vector<int>> graph;
  vector<int> match,used;
 
  BipartiteMatching(int n):graph(n),used(n, 0),match(n, -1){}
 
  void add_edge(int u,int v){
    graph[u].push_back(v);
    graph[v].push_back(u);
  }
 
  bool dfs(int now){
    used[now]=1;
    for(auto to:graph[now]){
      int to_match=match[to];
      if(to_match==-1||(!used[to_match]&&dfs(to_match))){
        match[now]=to;
        match[to]=now;
        return true;
      }
    }
    return false;
  }
 
  int bipartite_matching(){
    int res=0;
    for(int i=0;i<graph.size();i++){
      if(match[i]==-1){
        for(int j=0;j<graph.size();j++) used[j]=0;
        if(dfs(i)) res++;
      }
    }
    return res;
  }
};

int N,M;
int A[101][101];

set<int> se[101];
int B[101][101];
int C[101][101];

void solve(int col){
  BipartiteMatching BG(2*N+1);
  for(int i=0;i<N;i++){
    vector<int> used(N+1,0);
    for(auto val:se[i]){
      int t=(val-1)/M;
      if(used[t]) continue;
      used[t]=true;
      BG.add_edge(i,N+t);
    }
  }

  int m=BG.bipartite_matching();

  for(int i=0;i<N;i++){
    int tar=BG.match[i]-N;
    auto itr=se[i].lower_bound(tar*M+1);

    B[i][col]=*itr;
    se[i].erase(itr);
  }
  return;
}


int main(){
  cin>>N>>M;
  for(int i=0;i<N;i++){
    for(int j=0;j<M;j++){
      cin>>A[i][j];
      se[i].insert(A[i][j]);
    }
  }

  for(int i=0;i<M;i++) solve(i);

  for(int i=0;i<M;i++){
    vector<int> v;
    for(int j=0;j<N;j++) v.push_back(B[j][i]);
    sort(v.begin(),v.end());
    for(int j=0;j<N;j++) C[j][i]=v[j];
  }

  for(int i=0;i<N;i++){
    for(int j=0;j<M;j++) cout<<B[i][j]<<" ";
    cout<<endl;
  }

  for(int i=0;i<N;i++){
    for(int j=0;j<M;j++) cout<<C[i][j]<<" ";
    cout<<endl;
  }

  return 0;
}