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;

vector<int> Match(const vector<vector<int>> &g) {
    int n = g.size();
    vector<int> match(n, -1), used(n);

    function<bool(int)> Dfs = [&](int x) {
        for (int u : g[x]) {
            if (used[u]++) continue;
            if (match[u] == -1 || Dfs(match[u])) {
                match[u] = x;
                return true;
            }
        }
        return false;
    };

    for (int i = 0; i < n; ++i) {
        fill(used.begin(), used.end(), 0);
        Dfs(i);
    }
    return match;
}

int main() {
    int n, m; scanf("%d%d", &n, &m);
    vector<vector<int>> a(n, vector<int>(m));
    vector<vector<vector<int>>> cnt(n, vector<vector<int>>(n));
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            scanf("%d", &a[i][j]);
            a[i][j]--;
            cnt[i][a[i][j] / m].push_back(a[i][j]);
        }
    }
    vector<vector<int>> b(n, vector<int>(m, -1));
    for (int i = 0; i < m; ++i) {
        vector<vector<int>> g(n);
        for (int j = 0; j < n; ++j) {
            for (int k = 0; k < n; ++k) {
                if (!cnt[j][k].empty()) g[j].push_back(k);
            }
        }
        auto mk = Match(g);
        for (int j = 0; j < n; ++j) {
            b[mk[j]][i] = cnt[mk[j]][j].back();
            cnt[mk[j]][j].pop_back();
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) printf("%d ", b[i][j] + 1);
        puts("");
    }
    vector<vector<int>> c(n, vector<int>(m, -1));
    for (int i = 0; i < m; ++i) {
        vector<int> w(n, -1);
        for (int j = 0; j < n; ++j) w[b[j][i] / m] = b[j][i];
        for (int j = 0; j < n; ++j) c[j][i] = w[j];
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) printf("%d ", c[i][j] + 1);
        puts("");
    }
    return 0;
}