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 a[105][105],val[105],x[105][105],y[105][105],n,m,p[105*105],q[105*105],vis[105*105],v[105],to[105],ans;

bool dfs(int u){
	for (int i=1;i<=n;i++)
	    if (a[u][i]&&!v[i]){
	    	v[i]=true;
	    	if (to[i]==-1||dfs(to[i])){
	    		to[i]=u; val[i]=a[u][i];
	    		return true;
	    	}
	    }
    return false;
}
int main(){
    cin >> n >> m;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            cin >> x[i][j];
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            y[i][j]=(i-1)*m+j;
    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
            p[x[i][j]]=i,q[y[i][j]]=i;

    for (int t=1;t<=m;t++){
        memset(a,0,sizeof(a));
        for (int i=1;i<=n*m;i++) if (!vis[i]) a[p[i]][q[i]]=i;
        memset(to,-1,sizeof(to)); ans=0;
        for (int i=1;i<=n;i++){
            memset(v,0,sizeof(v));
            if (dfs(i)) ans++;
        }
        assert(ans==n);
        for (int i=1;i<=n;i++){
            y[i][t]=val[i];
            x[to[i]][t]=val[i];
            vis[val[i]]=1;
        }
    }
    for (int i=1;i<=n;i++){
        for (int j=1;j<m;j++) printf("%d ",x[i][j]);
        printf("%d\n",x[i][m]);
    }
    for (int i=1;i<=n;i++){
        for (int j=1;j<m;j++) printf("%d ",y[i][j]);
        printf("%d\n",y[i][m]);
    }
    return 0;
}