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=111;
int n,m;
int a[N][N];
bool f[N][N];
int cp[N];
bool vis[N];
bool dfs(int x){
	vis[x]=true;
	for(int i=0;i<n;i++){
		if(f[x][i]&&(cp[i]==-1||(!vis[cp[i]]&&dfs(cp[i])))){
			cp[i]=x;
			return true;
		}
	}
	return false;
}
void qaq(){
	memset(cp,-1,sizeof(cp));
	for(int i=0;i<n;i++){
		memset(vis,false,sizeof(vis));
		assert(dfs(i));
	}
}
void solve(){
	for(int k=0;k<m;k++){
		memset(f,0,sizeof(f));
		for(int i=0;i<n;i++){
			for(int j=k;j<m;j++){
				f[i][a[i][j]/m]=true;
			}
		}
		qaq();
		for(int i=0;i<n;i++){
			int r=cp[i];
			for(int j=k;j<m;j++){
				if(a[r][j]/m==i){
					swap(a[r][k],a[r][j]);
					break;
				}
			}
		}
	}
}
void rev(){
	for(int i=0;i<N;i++){
		for(int j=0;j<i;j++){
			swap(a[i][j],a[j][i]);
		}
	}
}
void print(){
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			printf("%d ",a[i][j]+1);
		}
		printf("\n");
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=0;i<n;i++){
		for(int j=0;j<m;j++){
			scanf("%d",&a[i][j]);
			a[i][j]--;
		}
	}
	solve();
	print();
	rev();
	for(int j=0;j<m;j++){
		sort(a[j],a[j]+n);
	}
	rev();
	print();
	return 0;
}