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;
 
const int N = 1 * 1000 + 17;
const int MOD = 1000 * 1000 * 1000 + 7;

struct tinder {
    int n, u;
    vector< vector<int> > g;
    vector<int> used, mt;

    tinder(int n) : n(n), u(1) {
        g.resize(n);
        used.resize(n, 0);
        mt.resize(n, -1);
    }

    void add_edge(int i, int j) {
        g[i].emplace_back(j);
    }
    
    bool dfs(int v) {
        if (used[v] == u)
            return false;

        used[v] = u;

        for (const auto& to : g[v])
            if (mt[to] == -1 || dfs(mt[to])) {
                mt[to] = v;
                return true;
            }

        return false;
    }

    vector<int> tyan() {
        for (int i = 0; i < n; ++i, ++u)
            dfs(i);

        assert(!count(mt.begin(), mt.end(), -1));
        vector<int> res(n);

        for (int i = 0; i < n; ++i)
            res[mt[i]] = i;

        return res;
    }
};

int n, m;
int a[N][N];

bool read() {
    if (!(cin >> n >> m))
        return false;

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            cin >> a[i][j], --a[i][j];

    return true;
}

void solve() {
    for (int j = 0; j < m; ++j) {
        tinder T(n);

        for (int i = 0; i < n; ++i)
            for (int j1 = j; j1 < m; ++j1)
                T.add_edge(i, a[i][j1] / m);
        
        auto mt = T.tyan();
        
        for (int i = 0; i < n; ++i)
            for (int j1 = j; j1 < m; ++j1)
                if (mt[i] == a[i][j1] / m) {
                    for (int j2 = j1; j2 > j; --j2)
                        swap(a[i][j2], a[i][j2 - 1]);

                    break;
                }
    }

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            cout << a[i][j] + 1 << " \n"[j == m - 1];

    for (int j = 0; j < m; ++j) {
        vector<int> sex;

        for (int i = 0; i < n; ++i)
            sex.emplace_back(a[i][j]);

        sort(sex.begin(), sex.end(), [] (int a, int b) {
            return a / m < b / m;
        });

        for (int i = 0; i < n; ++i)
            a[i][j] = sex[i];
    }

    for (int i = 0; i < n; ++i)
        for (int j = 0; j < m; ++j)
            cout << a[i][j] + 1 << " \n"[j == m - 1];
}
 
int32_t main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);

    while (read())
        solve();
 
    return 0;
}