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<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<vector>
#include<queue>
using namespace std;
const int N=205;
int n,m,cnt,a[N][N],b[N][N],c[N][N],d[N][N];
pair<int,int> bel[N];
vector<int> e[N][N];
namespace flow{
	int cnt,s,t,head[N],dep[N],to[N*N],nxt[N*N],dd[N*N];
	queue<int> q;
	void adde(int u,int v,int d){
		to[++cnt]=v;
		nxt[cnt]=head[u];
		dd[cnt]=d;
		head[u]=cnt;
		to[++cnt]=u;
		nxt[cnt]=head[v];
		dd[cnt]=0;
		head[v]=cnt;
	}
	void init(){
		cnt=1;
		s=0;
		t=2*n+1;
		memset(head,0,sizeof(head));
		for(int i=1;i<=n;i++){
			adde(s,i,1);
			adde(i+n,t,1);
			for(int j=1;j<=n;j++){
				if(e[i][j].size()){
					adde(i,j+n,1);
				}
			}
		}
	}
	bool bfs(){
		memset(dep,0,sizeof(dep));
		dep[s]=1;
		q.push(s);
		while(!q.empty()){
			int u=q.front(),v;
			q.pop();
			for(int i=head[u];i;i=nxt[i]){
				v=to[i];
				if(!dep[v]&&dd[i]){
					dep[v]=dep[u]+1;
					q.push(v);
				}
			}
		}
		return dep[t];
	}
	int dfs(int u,int f){
		if(u==t||!f){
			return f;
		}
		int v,tmp,res=0;
		for(int i=head[u];i&&f;i=nxt[i]){
			v=to[i];
			if(dep[v]==dep[u]+1&&(tmp=dfs(v,min(f,dd[i])))){
				dd[i]-=tmp;
				dd[i^1]+=tmp;
				f-=tmp;
				res+=tmp;
			}
		}
		if(!res){
			dep[u]=-1;
		}
		return res;
	}
	void maxflow(){
		while(bfs()){
			dfs(s,0x7fffffff);
		}
	}
}
int main(){
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			scanf("%d",&a[i][j]);
			bel[a[i][j]].first=i;
			b[i][j]=++cnt;
			bel[b[i][j]].second=i;
		}
	}
	for(int i=1;i<=n*m;i++){
		e[bel[i].first][bel[i].second].push_back(i);
	}
	for(int i=1;i<=m;i++){
		flow::init();
		flow::maxflow();
		for(int j=1;j<=n;j++){
			for(int k=flow::head[j];k;k=flow::nxt[k]){
				int l=flow::to[k];
				if(l&&!flow::dd[k]){
					l-=n;
					c[j][i]=d[l][i]=e[j][l].back();
					e[j][l].pop_back();
				}
			}
		}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			printf("%d ",c[i][j]);
		}
		puts("");
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++){
			printf("%d ",d[i][j]);
		}
		puts("");
	}
	return 0;
}