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 pb push_back
const int maxi = 1e6+10;
int a[110][110], b[110][110];
string s;
int n, m;
vector<int> v[maxi];
set<pair<int,int>> st;
int c[110][110];
int cnt[110][110];

int calc(int x)
{
    int val = x/m;
    if (x%m!=0) val++;
    return val;
}
int main()
{
    cin>>n>>m;

    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
        {
            scanf("%d",&a[i][j]);
            b[i][j] = a[i][j];
        }

     for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
     {
           int val = calc(b[i][j]);
           cnt[j][val]++;
     }


    int nula = 0;
    while(true)
    {
        for (int i=1;i<=n;i++)
        {
            for (int j=1;j<=m;j++)
                for (int k=j+1;k<=m;k++)
                if (cnt[j][calc(b[i][j])]>1 && cnt[k][calc(b[i][j])]==0)
                {
                   // cout<<"USAO :"<<i<<" "<<j<<" "<<k<<endl;
                    cnt[j][calc(b[i][j])]--;
                    cnt[k][calc(b[i][k])]--;
                    cnt[j][calc(b[i][k])]++;
                    cnt[k][calc(b[i][j])]++;
                    swap(b[i][j], b[i][k]);
                    nula++;

                }
        }

        if (!nula) break;
        nula = 0;
    }

    for (int i=1;i<=n;i++)
        for (int j=1;j<=m;j++)
    {
        int val = calc(b[i][j]);
        c[val][j] = b[i][j];
    }

    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=m;j++)
            printf("%d ",b[i][j]);
        printf("\n");
    }


    for (int i=1;i<=n;i++)
    {
        for (int j=1;j<=m;j++)
            printf("%d ",c[i][j]);
        printf("\n");
    }

    return 0;
}