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 inf=0x3f3f3f3f;
int n,m;
int sr[10010];
int A[110][110];
bool book[10010];
int B[110][110];
int C[110][110];

int s,t;
struct data{
    int to,nxt,c,id;
}mp[100100];
int head[1010],cnt=1;
void link(int x,int y,int c,int id)
{
    mp[++cnt].to=y;
    mp[cnt].nxt=head[x];
    mp[cnt].c=c;
    mp[cnt].id=id;
    head[x]=cnt;
}
void add(int x,int y,int c,int id)
{
    link(x,y,c,id);
    link(y,x,0,id);
}
int cur[1010];
int dis[1010];
queue<int>q;
bool BFS()
{
    memset(dis,0,sizeof(dis));dis[s]=1;
    q.push(s);
    int x;
    while(!q.empty())
    {
        x=q.front();q.pop();
        for(int i=head[x];i;i=mp[i].nxt)
            if(!dis[mp[i].to]&&mp[i].c&&!book[mp[i].id])
                dis[mp[i].to]=dis[x]+1,q.push(mp[i].to);
    }
    return dis[t];
}
int dfs(int x,int f)
{
    if(x==t)return f;
    int rf=0,fl;
    for(int& i=cur[x];i;i=mp[i].nxt)
        if(dis[mp[i].to]==dis[x]+1&&mp[i].c&&!book[mp[i].id])
        {
            fl=dfs(mp[i].to,min(f,mp[i].c));
            if(fl)
            {
                rf+=fl;
                f-=fl;
                mp[i].c-=fl;
                mp[i^1].c+=fl;
                if(!f)break;
            }
        }
    return rf;
}
void Dinic()
{
    int mf=0,fl;
    while(BFS())
    {
        memcpy(cur,head,sizeof(head));
        while(fl=dfs(s,inf))mf+=fl;
    }
}
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]);
            sr[m*(i-1)+j]=i;
        }
    s=n*2+1;t=s+1;
    for(int i=1;i<=n;++i)
        add(s,i,1,0),add(i+n,t,1,0);
    int tc=cnt;
    for(int i=1;i<=n;++i)
        for(int j=1;j<=m;++j)
            add(sr[A[i][j]],i+n,1,A[i][j]);
    int x;
    for(int j=1;j<=m;++j)
    {
        for(int i=2;i<=tc;i+=2) mp[i].c=1,mp[i^1].c=0;
        Dinic();
        for(x=1;x<=n;++x)
            for(int i=head[x];i;i=mp[i].nxt)
                if(!book[mp[i].id]&&mp[i].to>n&&!mp[i].c)
                {
                    book[mp[i].id]=1;
                    B[mp[i].to-n][j]=mp[i].id;
                    C[x][j]=mp[i].id;
                    break;
                }
    }
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=m;++j)printf("%d ",B[i][j]);
        puts("");
    }
    for(int i=1;i<=n;++i)
    {
        for(int j=1;j<=m;++j)printf("%d ",C[i][j]);
        puts("");
    }
    return 0;
}