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 Q=505,P=1<<20;
int las[Q],e[P],nn[P],c[P],inc;
int S,T;
int cnt[Q],dis[Q];
int Sap(int x,int fl)
{
    if(x==T)return fl;
    int had=0;
    for(int t=las[x];t;t=nn[t]){
        int y=e[t];
        if((!c[t])||dis[x]-1!=dis[y])continue;
        int res=Sap(y,min(c[t],fl-had));
        had+=res,c[t]-=res,c[t^1]+=res;
        if(had==fl||dis[S]>=T)return had;
    }
    --cnt[dis[x]];
    if(!cnt[dis[x]])dis[S]=T;
    cnt[++dis[x]]++;
    return had;
}
void iNS(int x,int y,int z)
{
    e[++inc]=y;
    nn[inc]=las[x];
    las[x]=inc;
    c[inc]=z;
}
void ins(int x,int y,int z)
{iNS(x,y,z),iNS(y,x,0);}
int a[Q][Q],b[Q][Q],cc[Q],uz[Q*Q];
int main()
{
    int n,m;
    scanf("%d%d",&n,&m);
    for(int i=1;i<=n;i++)
        for(int j=1;j<=m;j++)
            scanf("%d",&a[i][j]);
    for(int p=1;p<=m;p++){
        inc=1;
        S=(n<<1)+1,T=S+1;
        memset(las,0,sizeof(las));
        memset(cnt,0,sizeof(cnt));
        memset(dis,0,sizeof(dis));
        for(int i=1;i<=n;i++)
            for(int j=1;j<=m;j++)
                if(!uz[a[i][j]])
                    ins(i,(a[i][j]-1)/m+1+n,1);
        for(int i=1;i<=n;i++)
            ins(S,i,1),ins(i+n,T,1);
        int ch=0;
        while(dis[S]<T)ch+=Sap(S,(int)1e9);
        assert(ch==n);
        int now=1;
        for(int i=1;i<=n;i++){
            for(int j=1;j<=m;j++)
                if(!uz[a[i][j]]){
                    now+=2;
                    if(c[now]){
                        uz[a[i][j]]=1;
                        b[i][p]=a[i][j];
                    }
                }
        }
    }
    for(int i=1;i<=n;i++,puts(""))
        for(int j=1;j<=m;j++)
            printf("%d ",b[i][j]);
    for(int j=1;j<=m;j++){
        for(int i=1;i<=n;i++)
            cc[i]=b[i][j];
        sort(cc+1,cc+n+1);
        for(int i=1;i<=n;i++)
            b[i][j]=cc[i];
    }
    for(int i=1;i<=n;i++,puts(""))
        for(int j=1;j<=m;j++)
            printf("%d ",b[i][j]);
    return 0;
}