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>

#define MIN_INT -2147483648
#define MAX_INT 2147483647
#define MIN_LONG -9223372036854775808L
#define MAX_LONG 9223372036854775807L

#define long long long int

using namespace std;

// @author: pashka

int n, m;
bool q[100][100];
int r[100];
bool z[100];

bool dfs(int i) {
    if (z[i]) return false;
    z[i] = true;
    for (int j = 0; j < n; j++) {
        if (q[i][j]) {
            if (r[j] == -1 || dfs(r[j])) {
                r[j] = i;
                return true;
            }
        }
    }
    return false;
}

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

    cin >> n >> m;
    vector<vector<int>> a(n, vector<int>(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++) {
        for (int i = 0; i < n; i++) {
            r[i] = -1;
            for (int j = 0; j < n; j++) {
                q[i][j] = false;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = k; j < m; j++) {
                q[a[i][j] / m][i] = true;
            }
        }
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                z[j] = false;
            }
            if (!dfs(i)) {
                return 1;
            }
        }
        for (int i = 0; i < n; i++) {
            int x = r[i];
            for (int j = k; j < m; j++) {
                if (a[i][j] / m == x) {
                    swap(a[i][j], a[i][k]);
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m;j ++) {
            cout << a[i][j] + 1 << " ";
        }
        cout << "\n";
    }
    for (int k = 0; k < n; k++) {
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                if (a[i][j] / m == k) {
                    swap(a[i][j], a[k][j]);
                }
            }
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m;j ++) {
            cout << a[i][j] + 1 << " ";
        }
        cout << "\n";
    }

    return 0;
}