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;
#define ll long long
char buf[1<<20],*p1,*p2;
#define GC (p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<20,stdin),p1==p2)?0:*p1++)
template<class T> inline void read(T &n){
	char ch=GC;T w=1,x=0;
	while(!isdigit(ch)){if(ch=='-') w=-1;ch=GC;}
	while(isdigit(ch)){x=(x<<3)+(x<<1)+(ch^48);ch=GC;}
	n=x*w;
}
const int maxn=105;
const int maxm=10205;
int n,m,Map[maxn][maxn];
vector <int> col[maxn][maxn];
int last[maxn<<1],Next[maxm<<1],to[maxm<<1],w[maxm<<1],cnt=1;
inline void addedge(int x,int y,int z){
	Next[++cnt]=last[x],last[x]=cnt;
	to[cnt]=y,w[cnt]=z;
}
inline void add(int x,int y,int z){
	addedge(x,y,z),addedge(y,x,0);
}
int dep[maxn<<1],cur[maxn<<1],st,ed;
queue <int> Q;
bool bfs(){
	memset(dep,0,sizeof(dep));
	dep[st]=1,Q.push(st);
	while(!Q.empty()){
		int u=Q.front();Q.pop();
		for(int i=last[u];i;i=Next[i]){
			int v=to[i];
			if(w[i]&&!dep[v]){
				dep[v]=dep[u]+1;
				Q.push(v);
			}
		}
	}
	return dep[ed];
}
const int inf=0x3f3f3f3f;
int dfs(int u,int low){
	if(u==ed) return low;
	int add=0;
	for(int i=cur[u];i&&add<low;i=Next[i]){
		int v=to[i];cur[u]=i;
		if(w[i]&&dep[v]==dep[u]+1){
			int tmp=dfs(v,min(w[i],low-add));
			w[i]-=tmp,w[i^1]+=tmp;
			add+=tmp;
		}
	}
	return add;
}
int dinic(){
	int res=0;
	while(bfs()){
		memcpy(cur,last,sizeof(last));
		res+=dfs(st,inf);
	}
	return res;
}
int ans[maxn][maxn],bot[maxn];
int main(){
	read(n),read(m);
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){
			read(Map[i][j]);
			int bel=(Map[i][j]-1)/m+1;
			col[i][bel].push_back(Map[i][j]);
		}
 	ed=n*2+1;
	for(int now=1;now<=m;now++){
		memset(last,0,sizeof(last)),cnt=1;
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++)
				if(col[i][j].size()) add(i,j+n,1);
			add(st,i,1);
			add(i+n,ed,1);
		}
		assert(dinic()==n);
		for(int x=1;x<=n;x++)
			for(int i=last[x];i;i=Next[i]){
				int v=to[i];
				if(!w[i]){
					ans[x][now]=col[x][v-n].back();
					col[x][v-n].pop_back();
				}
			}
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++) printf("%d ",ans[i][j]);
		printf("\n");
	}
	for(int j=1;j<=m;j++){
		for(int i=1;i<=n;i++) bot[i]=ans[i][j];
		sort(bot+1,bot+n+1);
		for(int i=1;i<=n;i++) ans[i][j]=bot[i];
	}
	for(int i=1;i<=n;i++){
		for(int j=1;j<=m;j++) printf("%d ",ans[i][j]);
		printf("\n");
	}
	return 0;
}