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:
/*	
	Over ice, I'm freezing
	Beautiful eyes, deceiving
	We may die this evening
	Coughing, wheezing, bleeding...
*/	
#include <bits/stdc++.h>
using namespace std;

#pragma GCC optimize("O2,no-stack-protector,unroll-loops,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
//#pragma GCC optimize("Os")

#define F first
#define S second
#define pb push_back
#define SZ(x) (ll)(x.size())
#define all(x) x.begin(),x.end()
#define mp make_pair

typedef long long ll;
typedef long double ld;
typedef pair<ll,ll> pll;

//mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const ll maxn=1e2+10, maxm=2e3+10, lg=18, mod=1e9+7, inf=1e18;

ll n,m,A[maxn][maxn],B[maxn][maxn],C[maxn][maxn],tr[maxn][maxn];
bool us[maxn][maxn],dar[maxn],mk[maxn];
vector<ll> g[maxn];
ll mt[maxn];

bool dfs(ll v){
	if(mk[v]) return 0;
	mk[v]=1;
	for(auto u:g[v])if(mt[u]==-1 || dfs(mt[u])) return mt[u]=v,1;
	return 0;
}

int main(){
	ios_base::sync_with_stdio(0),cin.tie(0),cout.tie(0);
	
	cin>>n>>m;
	for(int r=0;r<n;r++)for(int c=0;c<m;c++) cin>>A[r][c], tr[r][(--A[r][c])/m]++;
	for(int c=0;c<m;c++){
		for(int i=0;i<n;i++) g[i].clear(), mt[i]=-1, dar[i]=0;
		for(int r=0;r<n;r++)for(int col=0;col<n;col++)if(tr[r][col]) g[col].pb(r);
		bool F=1;
		while(F){
			F=0;
			for(int col=0;col<n;col++)if(!dar[col]){
				memset(mk,0,sizeof(mk));
				F|=(dar[col]=dfs(col));
			}
		}
		for(int i=0;i<n;i++)for(int j=0;j<m;j++)if(!us[i][j] && mt[i]==A[i][j]/m){ us[i][j]=1; B[i][c]=A[i][j]; C[c][A[i][j]/m]=A[i][j]; tr[i][A[i][j]/m]--; break; }
	}
	for(int i=0;i<n;i++,cout<<endl)for(int j=0;j<m;cout<<1+B[i][j++]<<' ');
	for(int i=0;i<n;i++,cout<<endl)for(int j=0;j<m;cout<<1+C[j++][i]<<' ');
	
	return 0;
}