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;
typedef long long ll;
const int N = 100 + 10;
vector<int> g[N];
int vis[N],match[N];
bool dfs(int u){
    if(vis[u]) return false;
    vis[u] = 1;
    for(int v: g[u]) if(!match[v]||dfs(match[v])){
        match[v] = u;
        return true;
    }
    return false;
}
int a[N][N],b[N][N],c[N][N],ok[N][N],n,m;
int main(){
    cin>>n>>m;
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) cin>>a[i][j], ok[i][j] = 1;
    for(int col=1;col<=m;col++){
        for(int i=1;i<=n;i++) g[i].clear();
        for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) if(ok[i][j]){
            g[i].push_back((a[i][j]-1)/m+1);
        }
        memset(match,0,sizeof(match));
        for(int i=1;i<=n;i++) memset(vis,0,sizeof(vis)), dfs(i);
        for(int i=1;i<=n;i++){
            int x;
            for(int j=1;j<=n;j++) if(match[j]==i) x = j;
            for(int j=1;j<=m;j++) if(ok[i][j]){
                if((a[i][j]-1)/m+1==x){
                    b[i][col] = a[i][j];
                    ok[i][j] = 0;
                    break;
                }
            }
        }
    }
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) c[(b[i][j]-1)/m+1][j] = b[i][j];
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) printf("%d%c",b[i][j]," \n"[j==m]);
    for(int i=1;i<=n;i++) for(int j=1;j<=m;j++) printf("%d%c",c[i][j]," \n"[j==m]);
}