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>
#define MAXN 128*4
using namespace std;
int n,m;
vector<int>g[MAXN];
int a[MAXN][MAXN],b[MAXN][MAXN],us[MAXN][MAXN],used[MAXN],mt[MAXN];
void read()
{
    int i,j;
    cin>>n>>m;
    for(i=0; i<n; i++)
        for(j=0; j<m; j++)
            cin>>a[i][j];
}

bool try_kuhn (int v)
{
    if(used[v])  return false;
    used[v] = true;
    for (size_t i=0; i<g[v].size(); ++i)
    {
        int to = g[v][i];
        if (mt[to] == -1 || try_kuhn (mt[to]))
        {
            mt[to] = v;
            mt[v]=to;
            return true;
        }
    }
    return false;
}

void solve()
{
    int i,j,l;
    for(l=0; l<m; l++)
    {
        //cout<<"---------\n";
        for(i=0; i<n; i++)g[i].clear();
        for(i=0; i<n; i++)
        {
            for(j=0; j<m; j++)
            {
                if(!us[i][j])
                {
                    g[i].push_back(a[i][j]/m-(a[i][j]%m==0)+n);
                    g[a[i][j]/m-(a[i][j]%m==0)+n].push_back(i);
                    //cout<<i<<" "<<a[i][j]/m-(a[i][j]%m==0)+n<<endl;
                }
            }
        }
        //cout<<"---------\n";
        memset(mt,-1,sizeof(mt));
        for (int v=0; v<n; ++v)
        {

            memset(used,0,sizeof(used));
            try_kuhn (v);
        }

        for (int i=0; i<n; ++i)
        {
            //cout<<mt[i]<<endl;
            if (mt[i] != -1)
            {
                //printf ("%d %d\n", mt[i], i);
                for(j=0; j<m; j++)
                {
                    if(a[i][j]/m-(a[i][j]%m==0)+n==mt[i]&&!us[i][j])
                    {
                        us[i][j]=true;
                        ///cout<<" - > "<<i<<" "<<j<<endl;
                        b[i][l]=a[i][j];
                        break;
                    }
                }
            }
        }
    }
    for(i=0; i<n; i++)
    {
        for(j=0; j<m; j++)
        {
            cout<<b[i][j]<<" ";
        }
        cout<<endl;
    }
    vector<int>h;
    for(j=0; j<m; j++)
    {
        h.clear();
        for(i=0; i<n; i++)
        {
            h.push_back(b[i][j]);
        }
        sort(h.begin(),h.end());
        for(i=0; i<n; i++)
        {
            b[i][j]=h[i];
        }
    }
    for(i=0; i<n; i++)
    {
        for(j=0; j<m; j++)
        {
            cout<<b[i][j]<<" ";
        }
        cout<<endl;
    }
}

int main()
{
    read();
    solve();
    return 0;
}