Problem:
Score : 1100 points

Problem Statement

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:

First, for each of the N rows, rearrange the numbers written in it as you like.

Second, for each of the M columns, rearrange the numbers written in it as you like.

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.

Sample Input 1

3 2
2 6
4 3
1 5

Sample Output 1

2 6
4 3
5 1
2 1
4 3
5 6

Sample Input 2

3 4
1 4 7 10
2 5 8 11
3 6 9 12

Sample Output 2

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 C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
//#include <boost/multiprecision/cpp_int.hpp>
//typedef boost::multiprecision::cpp_int ll;
typedef long double dd;
//#define i_7 (ll)(1E9+7)
#define i_7 998244353
#define i_5 i_7-2
ll mod(ll a){
    ll c=a%i_7;
    if(c>=0)return c;
    return c+i_7;
}
typedef pair<ll,ll> l_l;
ll inf=(ll)1E17;
#define rep(i,l,r) for(ll i=l;i<=r;i++)
#define pb push_back
ll max(ll a,ll b){if(a<b)return b;else return a;}
ll min(ll a,ll b){if(a>b)return b;else return a;}
void Max(ll &pos,ll val){pos=max(pos,val);}//Max(dp[n],dp[n-1]);
void Min(ll &pos,ll val){pos=min(pos,val);}
void Add(ll &pos,ll val){pos=mod(pos+val);}
dd EPS=1E-9;
#define fastio ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
///////////////////////////////////////



int main(){fastio
    ll n,m;cin>>n>>m;
    ll a[n][m];rep(i,0,n-1)rep(j,0,m-1){cin>>a[i][j];a[i][j]--;}
    ll b[n][m];
    bool f[n][m];memset(f,true,sizeof(f));
    rep(j,0,m-1){
        b[0][j]=a[0][j];
        f[b[0][j]/m][j]=false;
    }
    rep(i,1,n-1){
        set<ll>s[m];
        rep(j,0,m-1){
            ll t=a[i][j]/m;
            rep(k,0,m-1){
                if(f[t][k])s[k].insert(a[i][j]);
            }
        }
        ll q=m;
        bool res[m];memset(res,false,sizeof(res));
        while(q--){
            bool flag=true;
            rep(j,0,m-1){
                if(res[j])continue;
                if(s[j].size()==1){
                    ll key=*s[j].begin();
                    b[i][j]=key;
                    rep(k,0,m-1){
                        s[k].erase(key);
                    }
                    flag=false;
                    res[j]=true;
                    f[key/m][j]=false;
                    break;
                }
            }
            if(flag){
                ll j=0;
                rep(qq,0,m-1)if(res[qq]==false){j=qq;break;}
                ll key=*s[j].begin();
                b[i][j]=key;
                rep(k,0,m-1){
                    s[k].erase(key);
                }
                flag=false;
                res[j]=true;
                f[key/m][j]=false;
            }
        }
    }
    rep(i,0,n-1){
        rep(j,0,m-1){
            cout<<b[i][j]+1;
            if(j!=m-1)cout<<" ";
        }cout<<endl;
    }
    rep(i,0,n-1){
        rep(k,0,m-1){
            rep(j,0,n-1){
                if(b[j][k]/m==i){
                    cout<<b[j][k]+1;
                    if(k!=m-1)cout<<" ";
                    break;
                }
            }
        }cout<<endl;
    }
    
    
    
    return 0;
}