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>
#define syosu(x) fixed<<setprecision(x)
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> P;
typedef pair<double,double> pdd;
typedef pair<ll,ll> pll;
typedef vector<int> vi;
typedef vector<vi> vvi;
typedef vector<double> vd;
typedef vector<vd> vvd;
typedef vector<ll> vl;
typedef vector<vl> vvl;
typedef vector<string> vs;
typedef vector<P> vp;
typedef vector<vp> vvp;
typedef vector<pll> vpll;
typedef pair<int,P> pip;
typedef vector<pip> vip;
const int inf=1<<30;
const ll INF=1ll<<60;
const double pi=acos(-1);
const double eps=1e-8;
const ll mod=1e9+7;
const int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};

int n,m;
vvi a,c,g;
vector<vvi> b;
vi match,used;

bool BM_dfs(int v){
	used[v]=1;
	for(auto u:g[v]){
		int w=match[u];
		if(w<0||!used[w]&&BM_dfs(w)){
			match[v]=u;
				match[u]=v;
			return 1;
		}
	}
	return 0;
}
int BM(){
	int res=0;
	match=vi(2*n,-1);
	for(int i=0;i<n;i++) if(match[i]<0){
		used=vi(2*n);
		if(BM_dfs(i)) res++;
	}
	return res;
}
void Change(int u){
	int v=match[u];
	match[u]=match[v]=-1;
	vi g_;
	for(auto w:g[u]) if(w!=v) g_.push_back(w);
	g[u]=g_;
	g_.clear();
	for(auto w:g[v]) if(w!=u) g_.push_back(w);
	g[v]=g_;
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	cin>>n>>m;
	a=c=vvi(n,vi(m));
	b=vector<vvi>(n,vvi(n));
	g=vvi(2*n);
	for(int i=0;i<n;i++) for(int j=0;j<m;j++){
		int t;
		cin>>t;
		t--;
		a[i][j]=t;
		b[i][t/m].push_back(t);
	}
	for(int i=0;i<n;i++) for(int j=0;j<n;j++) if(!b[i][j].empty()){
		g[i].push_back(n+j);
		g[n+j].push_back(i);
	}
	for(int i=0;i<m;i++){
		BM();
		for(int j=0;j<n;j++){
			c[j][i]=b[j][match[j]-n].back();
			b[j][match[j]-n].pop_back();
			if(b[j][match[j]-n].empty()) Change(j);
		}
	}
	for(int i=0;i<n;i++) for(int j=0;j<m;j++) cout<<c[i][j]+1<<" \n"[j==m-1];
	for(int i=0;i<m;i++){
		vi d(n);
		for(int j=0;j<n;j++) d[j]=c[j][i];
		sort(d.begin(),d.end());
		for(int j=0;j<n;j++) c[j][i]=d[j];
	}
	for(int i=0;i<n;i++) for(int j=0;j<m;j++) cout<<c[i][j]+1<<" \n"[j==m-1];
}