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;
using ll = long long;
using pi = pair<int, int>;
int n,m;
pi a[101][101];
int vis[102];
vector<int> v[101];
int pre[101];
int dfs(int x)
{
    if(vis[x]) return false;
    vis[x] = true;
    for(int y : v[x])
    {
        if(pre[y] == -1 || dfs(pre[y]))
        {
            pre[y] = x;
            return true;
        }
    }
    return false;
}
int conv(pi p)
{
    return p.first*m + p.second + 1;
}
int main()
{
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

    cin >> n >> m;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            int x;
            cin >> x;
            x--;
            a[i][j] = {x/m, x%m};
        }
    }
    for(int j=0;j<m;j++)
    {
        fill(pre, pre+n, -1);
        for(int i=0;i<n;i++)
            v[i].clear();
        for(int i=0;i<n;i++)
        {
            int num_cnt[101] = {};
            for(int k=j;k<m;k++)
            {
                num_cnt[a[i][k].first]++;
                if(num_cnt[a[i][k].first] == 1)
                {
                    v[i].push_back(a[i][k].first);
                }
            }
        }
        for(int i=0;i<n;i++)
        {
            fill(vis, vis+n, 0);
            dfs(i);
            ///cout << "dfs " << dfs(i) << '\n';
        }
        for(int i=0;i<n;i++)
        {
            //cout << "pre " << i << ' ' << pre[i] << '\n';
            for(int k=j;k<m;k++)
            {
                if(a[pre[i]][k].first == i)
                {
                    //cout << j << ' ' << k << '\n';
                    swap(a[pre[i]][k], a[pre[i]][j]);
                    break;
                }
            }
        }
        /*
        cout << '\n';
        for(int i=0;i<n;i++)
        {
            for(int j=0;j<m;j++)
            {
                cout << a[i][j].first << ' ';
            }
            cout << '\n';
        }*/
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            cout << conv(a[i][j]) << ' ';
        }
        cout << '\n';
    }
    for(int j=0;j<m;j++)
    {
        vector<pi> pv;
        for(int i=0;i<n;i++)
            pv.push_back(a[i][j]);
        sort(pv.begin(), pv.end());
        for(int i=0;i<n;i++)
            a[i][j] = pv[i];
    }
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            cout << conv(a[i][j]) << ' ';
        }
        cout << '\n';
    }
}