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 <iostream>
#include <algorithm>
#include <fstream>
#include <vector>
#include <deque>
#include <assert.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
#include <stdio.h>
#include <string.h>
#include <utility>
#include <math.h>
#include <bitset>
#include <iomanip>
#include <complex>
using namespace std;
//#define int long long
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long double ld;
typedef long long ll;
#define X first
#define Y second
#define all(o) o.begin(), o.end()
#define endl '\n'
#define IOS ios::sync_with_stdio(0), cin.tie(0)

int gcd(int x,int y){ return (!y ? x : gcd(y, x%y)); }
const int maxn = 100 + 5;
int a[maxn][maxn];
int n, m;
const int N = maxn * 2;
int g[N][N];
int mat[N];
void ad_e(int x,int y){
	y += n;
	g[x][y] = 1;
	g[y][x] = 1;
}
bool mark[N];
bool dfs(int v){ // can find aug path
	if(mark[v])
		return false;
	mark[v] = 1;
	for(int i=0; i<N; i++){
		if(g[v][i] == 0)
			continue;
		if(mat[i] == -1 || dfs(mat[i])){
			mat[i] = v;
			mat[v] = i;
			return true;
		}
	}
	return false;
}
int32_t main(){
	IOS;
	cin >> n >> m;
	for(int i=0; i<n; i++)
		for(int j=0; j<m; j++){
			cin >> a[i][j];
			a[i][j]--;
		}
	for(int col=0; col<m; col++){
		memset(mat, -1, sizeof mat);
		memset(g, 0, sizeof g);
		for(int i=0; i<n; i++){
			for(int j=col; j<m; j++){
				ad_e(i, a[i][j]/m);		
			}
		}
		int sz = 0;
		for(int i=0; i<n; i++){
			memset(mark, 0, sizeof mark);
			sz += dfs(i);
		}
		//cout << col << " " << sz << endl;
		for(int i=0; i<n; i++){
			int u = mat[i] - n;
			for(int j=col; j<m; j++){
				if(a[i][j]/m == u){
					swap(a[i][j], a[i][col]);
					break;
				}
			}
		}/*
		for(int i=0; i<n; i++){
			for(int j=0; j<m; j++){
				cout << a[i][j]/m << " ";
			}
			cout << endl;
		}*/
	}	
	for(int i=0; i<n; i++){
		for(int j=0; j<m; j++){
			cout << a[i][j] + 1 << " ";
		}
		cout << endl;
	}
	for(int i=0; i<m; i++){
		vi v;
		for(int j=0; j<n; j++)
			v.push_back(a[j][i]);
		sort(all(v));
		for(int j=0; j<n; j++)
			a[j][i] = v[j];
	}
	for(int i=0; i<n; i++){
		for(int j=0; j<m; j++){
			cout << a[i][j] + 1 << " ";
		}
		cout << endl;
	}
}