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 MAXN = 200;
int k = 1;
int p[MAXN], used[MAXN];
int in_m[MAXN];
vector<int> g[MAXN];
int arr[MAXN][MAXN];

bool dfs(int v) {
    // cout << k << " " << v << endl;
    used[v] = k;
    for (auto u : g[v]) {
        if (p[u] == -1) {
            p[u] = v;
            in_m[v] = 1;
            return true;
        }
        if (used[p[u]] != k && dfs(p[u])) {
            in_m[v] = 1;
            p[u] = v;
            return true;
        }
    }
    return false;
}

int main() {
    int n, m;
    cin >> n >> m;
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cin >> arr[i][j];
        }
    }
    for (int i = 0; i < m; ++i) {
        for (int j = 0; j < n; ++j) {
            g[j].clear();
            p[j] = -1;
            in_m[j] = 0;
        }
        for (int j = 0; j < n; ++j) {
            for (int k = i; k < m; ++k) {
                g[j].push_back((arr[j][k] + m - 1) / m - 1);
            }
        }
        for (int j = 0; j < n; ++j) {
            k++;
            dfs(j);
        }
        for (int j = 0; j < n; ++j) {
            int x = p[j];
            for (int k = i; k < m; ++k) {
                if ((arr[x][k] + m - 1) / m - 1 == j) {
                    swap(arr[x][k], arr[x][i]);
                    break;
                }
            }
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cout << arr[i][j] << " ";
        }
        cout << endl;
    }
    for (int i = 0; i < m; ++i) {
        vector<int> col;
        for (int j = 0; j < n; ++j) {
            col.push_back(arr[j][i]);
        }
        sort(col.begin(), col.end());
        for (int j = 0; j < n; ++j) {
            arr[j][i] = col[j];
        }
    }
    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < m; ++j) {
            cout << arr[i][j] << " ";
        }
        cout << endl;
    }
    return 0;
}