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 N = 1e2+10;
#define fr(i,low,up,st) for(int i = low;i <= up;i += st)
#define nfr(i,low,up,st) for(int i = low;i >= up;i -= st)
#define sz(a) a.siz()
#define clr(a) memset(a,0,sizeof a)

vector<int> e[N];
int y[N],vis[N],a[N][N];

bool solve(int x){
   fr(i,0,e[x].size()-1,1){
      int w = e[x][i];
      if(!vis[w]){
         vis[w] = 1;
         if(!y[w] || solve(y[w])){
            y[w] = x;
            return true;
         }
      }
   }
   return false;
}

int n,m;
int mk[N][N];
int b[N][N];
int main(){
   scanf("%d%d",&n,&m);
   fr(i,1,n,1)
      fr(j,1,m,1)
         scanf("%d",&a[i][j]);

   fr(j,1,m,1){
      fr(i,1,n,1){
         e[i].clear();
         fr(k,1,m,1){
            if(!mk[i][k])e[i].push_back((a[i][k]-1)/m+1);
         }
      }
      memset(y,0,sizeof y);
      fr(i,1,n,1){
         memset(vis,0,sizeof vis);
         solve(i);
      }
      fr(i,1,n,1){
         fr(k,1,m,1){
            if(mk[i][k])continue ;
            int cl = (a[i][k]-1)/m+1;
            if(y[cl] == i){
               mk[i][k] = j;
               break;
            }
         }
      }
   }
   fr(i,1,n,1)
      fr(k,1,m,1)
         b[i][mk[i][k]] = a[i][k];
   memcpy(a,b,sizeof b);

   fr(i,1,n,1){
      fr(j,1,m,1)
         printf("%d ",a[i][j]);
      puts("");
   }
   fr(i,1,n,1){
      fr(j,1,m,1)
         mk[(a[i][j]-1)/m+1][j] = a[i][j];
   }
   fr(i,1,n,1){
      fr(j,1,m,1)
         printf("%d ",mk[i][j]);
      puts("");
   }
   return 0;
}