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;

int n,m,a[N+9][N+9];

void into(){
  scanf("%d%d",&n,&m);
  for (int i=1;i<=n;++i)
	for (int j=1;j<=m;++j)
	  scanf("%d",&a[i][j]);
}

int id[N+9][N+9];

void Get_id(){
  for (int i=1;i<=n;++i)
	for (int j=1;j<=m;++j) id[i][j]=(a[i][j]-1)/m+1;
}

int e[N+9][N+9];

void Get_graph(){
  for (int i=1;i<=n;++i){
	for (int j=1;j<=n;++j) e[i][j]=0;
	for (int j=1;j<=m;++j)
	  if (id[i][j]) e[i][id[i][j]]=1;
  }
}

int par[N+9],vis[N+9];

bool Dfs_par(int k){
  for (int i=1;i<=n;++i)
	if (e[k][i]&&vis[i]<vis[0]){
	  vis[i]=vis[0];
	  if (!par[i]||Dfs_par(par[i])) {par[i]=k;return 1;}
	}
  return 0;
}

void KM(){
  for (int i=1;i<=n;++i) par[i]=vis[i]=0;
  for (int i=1;i<=n;++i) Dfs_par(vis[0]=i);
}

int ans[2][N+9][N+9];

void Get_ans0(){
  for (int i=1;i<=m;++i){
	Get_graph();
	KM();
    for (int j=1;j<=n;++j){
	  int t=par[j];
	  for (int k=1;k<=m;++k)
		if (id[t][k]==j) {id[t][k]=0;ans[0][t][i]=a[t][k];break;}
	}
  }
}

int tmp[N+9];

void Get_ans1(){
  for (int i=1;i<=m;++i){
	for (int j=1;j<=n;++j) tmp[j]=ans[0][j][i];
	sort(tmp+1,tmp+n+1);
	for (int j=1;j<=n;++j) ans[1][j][i]=tmp[j];
  }
}

void work(){
  Get_id();
  Get_ans0();
  Get_ans1();
}

void outo(){
  for (int i=0;i<2;++i)
	for (int j=1;j<=n;++j){
	  for (int k=1;k<=m;++k)
		printf("%d ",ans[i][j][k]);
	  puts("");
	}
}

int main(){
  into();
  work();
  outo();
  return 0;
}