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 MN = 105;
const int INF = 1e9;
int N,M,A[MN],B[MN],dist[MN],D[MN][MN],L[MN][MN];
bool used[MN],ch[MN][MN];
vector<int> G[MN],V[MN];

bool cmp(int A, int B)
{
    return (A-1)/M<(B-1)/M;
}

void bfs(){
    queue<int> Q;
    for(int i=0; i<N; i++){
        if(!used[i]){
            dist[i] = 0;
            Q.push(i);
        }
        else dist[i] = INF;
    }
    while(!Q.empty()){
        int a = Q.front();
        Q.pop();
        for(int b: G[a]){
            if(B[b] != -1 && dist[B[b]] == INF){
                dist[B[b]] = dist[a] + 1;
                Q.push(B[b]);
            }
        }
    }
}

bool dfs(int a){
    for(int b: G[a]){
        if(B[b]==-1||dist[B[b]]==dist[a]+1&&dfs(B[b])){
            used[a] = 1;
            A[a] = b;
            B[b] = a;
            return 1;
        }
    }
    return 0;
}

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 >> D[i][j];
        }
    }
    for(int t=0; t<M; t++){
        for(int i=0; i<N; i++){
            G[i].clear();
            for(int j=0; j<M; j++){
                if(ch[i][j]) continue;
                G[i].push_back((D[i][j]-1)/M);
            }
        }
        int match = 0;
        fill(A,A+MN,-1);
        fill(B,B+MN,-1);
        while(1){
            bfs();
            int flow = 0;
            for(int i=0; i<N; i++)
                if(!used[i] && dfs(i)) flow++;
            if(flow == 0) break;
            match += flow;
        }
        fill(used,used+MN,0);
        for(int i=0; i<N; i++){
            for(int j=0; j<M; j++){
                if(ch[i][j]) continue;
                if(A[i]==(D[i][j]-1)/M){
                    ch[i][j] = 1;
                    L[i][t] = D[i][j];
                    break;
                }
            }
        }
    }
    for(int i=0; i<N; i++){
        for(int j=0; j<M; j++){
            cout << L[i][j] << ' ';
            V[j].push_back(L[i][j]);
        }
        cout << '\n';
    }
    for(int j=0; j<M; j++) sort(V[j].begin(),V[j].end(),cmp);
    for(int i=0; i<N; i++){
        for(int j=0; j<M; j++){
            cout << V[j][i] << ' ';
        }
        cout << '\n';
    }
}