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 MN 100
#define S 0
#define T 201
#define INF 2000000000
using namespace std;
inline int read()
{
	int x=0,f=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
	return x*f;
}
bool b[MN+5][MN+5];
int n,m,a[MN+5][MN+5],top,A[MN+5][MN+5],cnt=1,head[T+5],c[T+5],q[T+5],d[T+5];
struct edge{int to,next,w;}e[MN*MN*5+5];
inline void ins(int f,int t,int w)
{
    e[++cnt]=(edge){t,head[f],w};head[f]=cnt;
    e[++cnt]=(edge){f,head[t],0};head[t]=cnt;
}
int dfs(int x,int f)
{
    if(x==T) return f;int used=0;
    for(int&i=c[x];i;i=e[i].next)
        if(e[i].w&&d[e[i].to]==d[x]+1)
        {
            int w=dfs(e[i].to,min(f-used,e[i].w));
            used+=w;e[i].w-=w;e[i^1].w+=w;
            if(used==f) return f;
        }
    return used;
}
bool bfs()
{
    memset(d,0,sizeof(d));d[q[top=1]=S]=1;
    for(int i=1,j;i<=top;++i)
        for(j=c[q[i]]=head[q[i]];j;j=e[j].next)
            if(e[j].w&&!d[e[j].to]) d[q[++top]=e[j].to]=d[q[i]]+1;
    return d[T];
}
void Solve(int tms)
{
    memset(head,0,sizeof(head));cnt=1;
    for(int i=1;i<=n;++i) ins(S,i,1);
    for(int i=1;i<=n;++i) ins(i+n,T,1);
    for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) 
        if(!b[i][j]) ins(i,(a[i][j]+m-1)/m+n,1);
    int rr=0;
    while(bfs()) rr+=dfs(S,INF);
    for(int i=1;i<=n;++i)
        for(int j=head[i];j;j=e[j].next)
            if(e[j].to!=S&&!e[j].w)
            {
                for(int k=1;k<=m;++k)
                    if(!b[i][k]&&(a[i][k]+m-1)/m==e[j].to-n)
                    {
                        A[i][tms]=a[i][k];
                        b[i][k]=1;
                        break;
                    }
                break;
            }
}
int main()
{
    n=read();m=read();
    for(int i=1;i<=n;++i) for(int j=1;j<=m;++j) a[i][j]=read();
    for(int i=1;i<=m;++i) Solve(i);
    for(int i=1;i<=n;++i,puts(""))
        for(int j=1;j<=m;++j) printf("%d ",A[i][j]);
    for(int i=1;i<=m;++i)
    {
        for(int j=1;j<=n;++j) for(int k=j+1;k<=n;++k)
            if(A[j][i]>A[k][i]) swap(A[j][i],A[k][i]);
    }
    for(int i=1;i<=n;++i,puts(""))
        for(int j=1;j<=m;++j) printf("%d ",A[i][j]);
    return 0;
}