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<vector>
#include<bitset>
using namespace std;
int n, m, i, j, jj, ok, ii;
int a[105][105], x[105], y[105];
vector<int> v[105];
bitset<105> viz;
int cuplaj(int nod){
    if(viz[nod] == 1){
        return 0;
    }
    viz[nod] = 1;
    for(int i = 0; i < v[nod].size(); i++){
        int vecin = v[nod][i];
        if(y[vecin] == 0){
            x[nod] = vecin;
            y[vecin] = nod;
            return 1;
        }
    }
    for(int i = 0; i < v[nod].size(); i++){
        int vecin = v[nod][i];
        if(cuplaj(y[vecin]) == 1){
            x[nod] = vecin;
            y[vecin] = nod;
            return 1;
        }
    }
    return 0;
}
int main(){
    cin>> n >> m;
    for(i = 1; i <= n; i++){
        for(j = 1; j <= m; j++){
            cin>> a[i][j];
            a[i][j]--;
        }
    }
    for(j = 1; j <= m; j++){
        for(i = 1; i <= n; i++){
            v[i].clear();
            x[i] = y[i] = 0;
        }
        for(i = 1; i <= n; i++){
            for(jj = j; jj <= m; jj++){
                v[i].push_back(a[i][jj] / m + 1);
            }
        }
        do{
            ok = 0;
            viz.reset();
            for(i = 1; i <= n; i++){
                if(x[i] == 0 && cuplaj(i) ){
                    ok = 1;
                }
            }
        }while(ok == 1);

        for(i = 1; i <= n; i++){
            for(jj = j; jj <= m; jj++){
                if(a[i][jj] / m + 1 == x[i]){
                    swap(a[i][jj], a[i][j]);
                    break;
                }
            }
        }
    }
    for(i = 1; i <= n; i++){
        for(j = 1; j <= m; j++){
            cout<< a[i][j] + 1 <<" ";
        }
        cout<<"\n";
    }
    for(j = 1; j <= m; j++){
        for(i = 1; i <= n; i++){
            for(ii = i; ii <= n; ii++){
                if(a[ii][j] / m + 1 == i){
                    swap(a[ii][j], a[i][j]);
                    break;
                }
            }
        }
    }
    for(i = 1; i <= n; i++){
        for(j = 1; j <= m; j++){
            cout<< a[i][j] + 1 <<" ";
        }
        cout<<"\n";
    }
}