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 kN = 1e2+7;

int n, m;
int a[kN][kN], sum[kN][kN], sort_a[kN], vis[kN], co[kN];
// sum[i][j] the number of group j in row i
// co[i] the group of row i
// vis[i] has group i been visited

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

bool check(int cur)
{
    for (int i = 1; i <= n; ++i) {
        if (!sum[cur][i] || vis[i]) continue;
        vis[i] = 1;
        if (!co[i] || check(co[i])) {
            co[i] = cur;
            return true;
        }
    }
    return false;
}

int main()
{
    cin >> n >> m;
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= m; ++j) {
            cin >> a[i][j];
            int group = (a[i][j]-1)/m+1;
            ++sum[i][group];
        }
    }
    // phase 1
    for (int j = 1; j <= m; ++j) {
        memset(co, 0, sizeof co);
        for (int i = 1; i <= n; ++i) {
            memset(vis, 0, sizeof vis);
            check(i);
        }
        for (int g = 1; g <= n; ++g) {
            int i = co[g];
            --sum[i][g];
            for (int k = j; k <= m; ++k) {
                int group = (a[i][k]-1)/m+1;
                if (group == g) {
                    swap(a[i][j], a[i][k]);
                    break;
                }
            }
        }
    }
    print();
    // phase 2
    for (int j = 1; j <= m; ++j) {
        for (int i = 1; i <= n; ++i) {
            sort_a[i] = a[i][j];
        }
        sort(sort_a+1, sort_a+n+1);
        for (int i = 1; i <= n; ++i) {
            a[i][j] = sort_a[i];
        }
    }
    print();
    return 0;
}