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 = 105;
typedef vector<int> vi;
typedef pair<int,int> ii;
typedef vector<ii> vii;
int N,M;
vi match, vis, id;
vii adjlist[MAXN];
int used[MAXN*MAXN];
int ans[MAXN][MAXN];
int dfs(int u){
    if (vis[u]) return 0;
    vis[u] = 1;
    for (auto v : adjlist[u]){
        if (used[v.second] == 0 && (match[v.first] == -1 || dfs(match[v.first]))){
            match[v.first] = u;
            id[u] = v.second;
            match[u] = v.first;
            return 1;
        }
    }
    return 0;
}
int main(){
    scanf("%d%d",&N,&M);
    int L = N;
    for (int i = 0; i < N*M; i++){
        int x;
        scanf("%d",&x);
        //printf("%d %d %d\n",(x-1)/M,N+i/M,x);
        adjlist[(x-1)/M].push_back({N+i/M,x});
    }
    int MCBM = 0;
    for (int k = 0; k < M; k++){
        match.assign(2*N,-1);
        id.assign(2*N,-1);
        for (int i = 0; i < N; i++){
            vii random;
            for (auto x : adjlist[i]){
                if (match[x.first] == -1 && used[x.second] == 0){
                    random.push_back(x);
                }
            }
            if (random.size()){
                MCBM++;
                ii lucky = random[rand()%random.size()];
                //printf("will use edge %d %d %d\n",i,lucky.first,lucky.second);
                match[lucky.first] = i;
                id[i] = lucky.second;
                match[i] = lucky.first;
            }
            random.clear();
        }
        for (int i = 0; i < L; i++){
            if (match[i] == -1){
                vis.assign(2*N,0);
                dfs(i);
            }
        }
        for (int i = 0; i < L; i++){
            ans[match[i]-N][k] = id[i];
            used[id[i]] = 1;
        }
    }
    for (int i = 0; i < N; i++){
        for (int j = 0; j < M; j++){
            printf("%d ",ans[i][j]);
        }
        printf("\n");
    }
    for (int j = 0; j < M; j++){
        vector<int> v;
        for (int i = 0; i < N; i++){
            v.push_back(ans[i][j]);
        }
        sort(v.begin(),v.end());
        for (int i = 0; i < N; i++){
            ans[i][j] = v[i];
        }
    }
    for (int i = 0; i < N; i++){
        for (int j = 0; j < M; j++){
            printf("%d ",ans[i][j]);
        }
        printf("\n");
    }


}