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>

using namespace std;

class BipartiteMatch {
public:
    explicit BipartiteMatch(int N, int M) :
        N(N), M(M), g(N+M), match(N+M, -1), visit(N+M, 0)
    {}
public:
    void addEdge(int a, int b){
        g[a].push_back(b+N);
        g[b+N].push_back(a);
    }
    int maximumMatch(){
        int res = 0;
        fill(match.begin(), match.end(), -1);
        for(int i=0;i<N;i++){
            fill(visit.begin(), visit.end(), 0);
            if(search_(i)) ++res;
        }
        return res;
    }
    int getMatch(int t) const { return t < N ? match[t] - N : match[t]; }
private:
    bool search_(int u){
        if(u < 0) return true;
        for(auto next : g[u]){
            if(visit[next]) continue;
            visit[next] = 1;
            if(search_(match[next])){
                match[u] = next, match[next] = u;
                return true;
            }
        }
        return false;
    }
private:
    const int N, M;
    vector<vector<int>> g;
    vector<int> match;
    vector<int> visit;
};

int main(){
    int N, M; cin >> N >> M;
    vector<vector<int>> A(N, vector<int>(M));
    vector<vector<vector<int>>> val(N, vector<vector<int>>(N));
    for(int i=0;i<N;i++){
        for(int j=0;j<M;j++){
            int a; cin >> a;
            val[i][(a-1)/M].push_back(a);
        }
    }
    vector<vector<int>> B(N);
    vector<vector<int>> C(N);
    for(int i=0;i<M;i++){
        BipartiteMatch bm(N, N);
        for(int j=0;j<N;j++){
            for(int k=0;k<N;k++){
                if(!val[j][k].empty()) bm.addEdge(j, k);
            }
        }
        bm.maximumMatch();
        for(int j=0;j<N;j++){
            int m = bm.getMatch(j);
            int b = val[j][m].back();
            val[j][m].pop_back();
            B[j].push_back(b);
            C[(b-1)/M].push_back(b);
        }
    }
    for(auto& v : B){
        cout << v[0];
        for(int i=1;i<v.size();i++) cout << " " << v[i];
        cout << endl;
    }
    for(auto& v : C){
        cout << v[0];
        for(int i=1;i<v.size();i++) cout << " " << v[i];
        cout << endl;
    }
}