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<cstring>
using namespace std;
const int N=105;
int n,m,x,B[N][N],lf[N],C[N][N],d[N],f[N];
bool vis[N],e[N][N];
vector<int>vec[N][N];
bool dfs(int v){
    for(int i=1;i<=n;++i)
    if(!vis[i]&&e[v][i]){
        vis[i]=1;
        if(!lf[i]||dfs(lf[i])){
            lf[i]=v;
            return 1;
        }
    }
    return 0;
}
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;++i){
        for(int j=1;j<=m;++j)
        cin>>x,vec[i][(x-1)/m+1].push_back(x);
    }
    for(int i=1;i<=m;++i){
        memset(e,0,sizeof e);
        for(int j=1;j<=n;++j)
        for(int k=1;k<=n;++k)e[j][k]=!vec[j][k].empty();
        memset(lf,0,sizeof lf);
        for(int j=1;j<=n;++j){
            memset(vis,0,sizeof vis);
            dfs(j);
        }
        for(int j=1;j<=n;++j){
            d[lf[j]]=f[j]=vec[lf[j]][j].back();
            vec[lf[j]][j].pop_back();
        }
        for(int j=1;j<=n;++j)B[j][i]=d[j],C[j][i]=f[j];
    }
    for(int i=1;i<=n;++i)
    for(int j=1;j<=m;++j)
    cout<<B[i][j]<<" \n"[j==m];
    for(int i=1;i<=n;++i)
    for(int j=1;j<=m;++j)
    cout<<C[i][j]<<" \n"[j==m];
    return 0;
}