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>
#define rep(xx,yy) for(int xx=0;xx<yy;xx++)
const int N=203;
using namespace std;
int n,m,x,S,T,t,q[N],b[N][N],d[N];
stack<int>a[N][N];
bool e[N][N];
inline bool bfs(){
	rep(i,T+1)d[i]=0;
	d[q[0]=S]=t=1;
	rep(h,t){
		x=q[h];
		rep(i,T+1)
		if(e[x][i]&&!d[i])d[q[t++]=i]=d[x]+1;
	}
	return d[T];
}
bool dfs(int x,int f){
	if(x==T)return f;
	bool c,re=0;
	rep(i,T+1)
	if(f&&e[x][i]&&d[i]==d[x]+1)
	c=dfs(i,1),re|=c,f-=c,
	e[x][i]^=c,e[i][x]^=c;
	return re;
}
int main(){
	scanf("%d%d",&n,&m);S=n<<1,T=S|1;
	rep(i,n)rep(j,m)
	scanf("%d",&x),a[i][(x-1)/m].push(x);
	rep(i,m){
		memset(e,0,sizeof e);
		rep(j,n){
			e[S][j]=e[n+j][T]=1;
			rep(k,n)e[j][n+k]=a[j][k].size();
		}
		for(;bfs();dfs(S,N));
		rep(j,n)rep(k,n)
		if(e[k+n][j])b[i][j]=a[j][k].top(),a[j][k].pop();
	}
	rep(i,n)rep(j,m)
	printf("%d%c",b[j][i],j<m-1?' ':'\n');
	rep(i,m)sort(b[i],b[i]+n);
	rep(i,n)rep(j,m)
	printf("%d%c",b[j][i],j<m-1?' ':'\n');
}