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 maxn=105;

int n,m;
int a[maxn][maxn],b[maxn][maxn];
int ecnt,dep[1005],cur[1005],head[1005];
struct Edge{int v,next,flow;}edge[100005];
stack<int> c[maxn][maxn];

inline int get_col(int x){return (x-1)/m+1;}
inline void add_edge(int u,int v,int flow){
	edge[ecnt]=(Edge){v,head[u],flow},head[u]=ecnt++;
	edge[ecnt]=(Edge){u,head[v],0},head[v]=ecnt++;
}

bool bfs(int s,int t){
	memset(dep+1,0,4*t);
	queue<int> Q;dep[s]=1,Q.push(s);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int i=head[u],v;~i;i=edge[i].next){
			v=edge[i].v;
			if(edge[i].flow&&!dep[v])
				dep[v]=dep[u]+1,Q.push(v);
		}
	}
	return dep[t];
}

int dfs(int u,int flow,const int &t){
	if(u==t)return flow;
	int used=0,rlow;
	for(int v,&i=cur[u];~i;i=edge[i].next){
		v=edge[i].v;
		if(edge[i].flow&&dep[v]==dep[u]+1){
			rlow=dfs(v,min(edge[i].flow,flow-used),t);
			if(rlow)edge[i].flow-=rlow,edge[i^1].flow+=rlow,used+=rlow;
			if(used==flow)break;
		}
	}
	if(!used)dep[u]=-1;
	return used;
}

int main(){
	scanf("%d%d",&n,&m);
	memset(head+1,0xff,4*(n*2+2));
	for(int i=1;i<=n;i++)
		for(int j=1,x;j<=m;j++){
			scanf("%d",&a[i][j]),x=get_col(a[i][j]);
			c[i][x].push(a[i][j]),add_edge(i,n+x,1);
		}
	int s=n*2+1,t=s+1;
	for(int id=1;id<=m;id++){
		for(int i=1;i<=n;i++)
			add_edge(s,i,1),add_edge(i+n,t,1);
		while(bfs(s,t)){
			memcpy(cur+1,head+1,4*t);
			dfs(s,0x3f3f3f3f,t);
		}
		for(int i=n+1;i<=2*n;i++)
			for(int j=head[i],v;~j;j=edge[j].next)
				if(edge[j].flow){
					edge[j].flow=0,v=edge[j].v;
					b[v][id]=c[v][i-n].top(),c[v][i-n].pop();
				}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++)
			cout<<b[i][j]<<' ',a[get_col(b[i][j])][j]=b[i][j];
		cout<<'\n';
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++)cout<<a[i][j]<<' ';
		cout<<'\n';
	}
	return 0;
}