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<algorithm>
#include<cstring>
#include<vector>
using namespace std;
const int N=205;
struct edge{int to,nxt,w;}a[N*N];
int n,m,S,T,head[N],cnt,dep[N],q[N],hd,tl,cur[N],ans[N][N];
vector<int>vec[N][N];
void link(int u,int v){
	a[++cnt]=(edge){v,head[u],1};head[u]=cnt;
	a[++cnt]=(edge){u,head[v],0};head[v]=cnt;
}
bool bfs(){
	memset(dep,0,sizeof(dep));
	dep[q[hd=tl=1]=S]=1;
	while(hd<=tl)
		for(int u=q[hd++],e=head[u];e;e=a[e].nxt)
			if(a[e].w&&!dep[a[e].to]){
				dep[q[++tl]=a[e].to]=dep[u]+1;
				if(a[e].to==T)return true;
			}
	return false;
}
int dfs(int u,int f){
	if(u==T)return f;int res=0;
	for(int &e=cur[u];e;e=a[e].nxt)
		if(a[e].w&&dep[a[e].to]==dep[u]+1){
			int d=dfs(a[e].to,min(a[e].w,f));
			a[e].w-=d;a[e^1].w+=d;f-=d;res+=d;
			if(!f)break;
		}
	if(!res)dep[u]=0;return res;
}
void dinic(){
	while(bfs())memcpy(cur,head,sizeof(cur)),dfs(S,1<<30);
}
int main(){
	scanf("%d%d",&n,&m);S=n+n;T=n+n+1;
	for(int i=0,x;i<n;++i)
		for(int j=0;j<m;++j){
			scanf("%d",&x);--x;
			vec[i][x/m].push_back(x);
		}
	for(int i=0;i<m;++i){
		memset(head,0,sizeof(head));cnt=1;
		for(int j=0;j<n;++j){
			for(int k=0;k<n;++k)
				if(vec[j][k].size())
					link(j,k+n);
			link(S,j);link(j+n,T);
		}
		dinic();
		for(int j=0;j<n;++j){
			for(int e=head[j];e;e=a[e].nxt)
				if(!a[e].w){
					int k=a[e].to-n;
					ans[i][j]=vec[j][k].back();
					vec[j][k].pop_back();
				}
		}
	}
	for(int i=0;i<n;++i,puts(""))
		for(int j=0;j<m;++j)
			printf("%d ",ans[j][i]+1);
	for(int i=0;i<m;++i)
		sort(ans[i],ans[i]+n);
	for(int i=0;i<n;++i,puts(""))
		for(int j=0;j<m;++j)
			printf("%d ",ans[j][i]+1);
	return 0;
}