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 <algorithm>
#include <cassert>

using namespace std;

template<int n>
struct BG {
    vector<int> adj[n];
    bool vis[n];
    int match[n];

    void clear() {
        for (int u = 0; u < n; ++u) {
            adj[u].clear();
        }
    }

    void add_edge(int x, int y) {
        adj[x].push_back(y);
    }

    bool dfs(int x) {
        for (auto y : adj[x]) {
            if (vis[y]) continue;
            vis[y] = true;
            if (!~match[y] || dfs(match[y])) {
                match[y] = x;
                return true;
            }
        }
        return false;
    }

    int max_match() {
        int cnt = 0;
        for (int u = 0; u < n; ++u) match[u] = -1;
        for (int u = 0; u < n; ++u) {
            for (int v = 0; v < n; ++v) vis[v] = false;
            cnt += dfs(u);
        }
        return cnt;
    }
};

const int maxn = 100;

int n, m;
BG<maxn> bg;
int a[maxn][maxn];

void print() {
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cout << a[i][j] + 1 << " ";
        }
        cout << endl;
    }
}

int main() {
    ios::sync_with_stdio(false);

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

    for (int k = 0; k < m; ++k) {
        bg.clear();
        for (int i = 0; i < n; ++i) {
            for (int j = k; j < m; ++j) {
                bg.add_edge(a[i][j] / m, i);
            }
        }
        assert(bg.max_match() == n);
        for (int i = 0; i < n; ++i) {
            for (int j = k; j < m; ++j) {
                if (bg.match[i] == a[i][j] / m) {
                    swap(a[i][k], a[i][j]);
                }
            }
        }
    }

    print();

    for (int j = 0; j < m; ++j) {
        for (int i = 0; i < n; ++i) {
            for (int k = i; k < n; ++k) {
                if (a[k][j] / m == i) {
                    swap(a[k][j], a[i][j]);
                    break;
                }
            }
        }
    }

    print();

    return 0;
}