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;


void solve() {
    int n, m;
    cin >> n >> m;
    auto club = [&](int x){
        return x/m;
    };
    vector<int> a[101][101];
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            int x; cin >> x; x--;
            a[i][club(x)].emplace_back(x);
        }
    }
    int b[101][101];
    for (int j = 0; j < m; j++) {
        vector<vector<int>> g(n);
        for (int i = 0; i < n; i++) {
            for (int c = 0; c < n; c++) {
                if (!a[i][c].empty())
                    g[i].emplace_back(c);
            }
        }
        vector<bool> trk(n, false);
        vector<int> left(n, -1);
        function<bool(int)> dfs = [&](int l){
            if (trk[l]) return false;
            trk[l] = true;
            for (int r: g[l]) {
                if (left[r] == -1 || dfs(left[r])) {
                    left[r] = l;
                    return true;
                }
            }
            return false;
        };
        int cnt = 0;
        for (int i = 0; i < n; i++) {
            fill(trk.begin(), trk.end(), false);
            if (dfs(i)) cnt++;
        }
        assert(cnt == n);
        for (int c = 0; c < n; c++) {
            int i = left[c];
            b[i][j] = a[i][c].back(); a[i][c].pop_back();
        }
    }
    int c[101][101];
    for (int j = 0; j < m; j++) {
        for (int i = 0; i < n; i++) {
            c[j][i] = b[i][j];
        }
        sort(c[j], c[j]+n, [&](int x, int y){return club(x) < club(y);});
    }

    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cout << b[i][j]+1 << ' ';
        }cout << "\n";
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < m; j++) {
            cout << c[j][i]+1 << ' ';
        }cout << "\n";
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    cout << endl;
}