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 Maxn 307
int n,m;
int a[Maxn][Maxn],que[Maxn];
bool vis[Maxn];
vector<int> e[2*Maxn];
int l[Maxn];
void print()
{
    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=m;j++)
            printf("%d ",a[i][j]);
        printf("\n");
    }
}
void insert(int x,int y)
{
    e[x].push_back(n+y);
    e[n+y].push_back(x);
}
int find(int u)
{
    for (int i=0;i<(int)e[u].size();i++)
    {
        int v=e[u][i];
        if (!vis[v])
        {
            vis[v]=true;
            if (l[v]==0||find(l[v]))
            {
                l[v]=u;
                return true;
            }
        }
    }
    return false;
}
int hungary()
{
    int ans=0;
    memset(l,0,sizeof(l));
    for (int i=n+1;i<=2*n;i++)
    {
        memset(vis,false,sizeof(vis));
        if (find(i)) ++ans;
    }
    return ans;
}
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]);
    for (int i=1;i<=m;i++)
    {
        memset(e,0,sizeof(e));
        for (int j=1;j<=n;j++)
        {
            memset(vis,false,sizeof(vis));
            for (int k=i;k<=m;k++)
            {
                int tmp=(a[j][k]+m-1)/m;
                if (!vis[tmp])
                {
                    vis[tmp]=true;
                    insert(j,tmp);
                }
            }
        }
        hungary();
        for (int j=1;j<=n;j++)
        {
            int pos=i;
            for (int k=i;k<=m;k++)
            {
                int tmp=(a[j][k]+m-1)/m;
                if (tmp==l[j]-n)
                {
                    pos=k;
                    break;
                }
            }
            if (pos!=i) swap(a[j][pos],a[j][i]);
        }
    }
    print();
    for (int i=1;i<=m;i++)
    {
        for (int j=1;j<=n;j++)
            que[j]=a[j][i];
        sort(que+1,que+n+1);
        for (int j=1;j<=n;j++)
            a[j][i]=que[j];
    }
    print();
    return 0;
}