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;

int n, m, a[111][111], b[111][111], b2[111][111];

int G[111][111], pre[111], usd[111];

int fin(int x){
    for (int y=0;y<n;++y){
        if (usd[y]||!G[x][y]) continue;
        usd[y]=1;
        if (pre[y]==-1||fin(pre[y])){
            pre[y]=x;
            return 1;
        }
    }
    return 0;
}

int main(){
    cin>>n>>m;
    for (int i=0;i<n;++i)
        for (int j=0;j<m;++j){
            cin>>a[i][j]; --a[i][j];
        }
    for (int c=0;c<m;++c){
        memset(G,0,sizeof G);
        for (int i=0;i<n;++i)
            for (int j=0;j<m;++j)
                if (~a[i][j]) G[i][a[i][j]/m]=1;
        memset(pre,-1,sizeof pre);
        for (int i=0;i<n;++i){
            memset(usd,0,sizeof usd);
            fin(i);
        }
        for (int i=0;i<n;++i){
            int pi=pre[i];
            for (int j=0;j<m;++j)
                if (~a[pi][j]&&a[pi][j]/m==i){
                    b[pi][c]=b2[i][c]=a[pi][j];
                    a[pi][j]=-1;
                    break;
                }
        }
    }
    for (int i=0;i<n;++i){
        for (int j=0;j<m;++j)
            cout<<b[i][j]+1<<' ';
        cout<<endl;
    }
    for (int i=0;i<n;++i){
        for (int j=0;j<m;++j)
            cout<<b2[i][j]+1<<' ';
        cout<<endl;
    }
}