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;
const int MX = 105;
#define rep(i,n) for(int i=0; i<(n); i++)
int n, m, a, pr[MX], w, b[MX][MX];
set<int> st[MX];
bool used[MX];
bool dfs(int v){
	used[v] = true;
	for(int it: st[v]) if((w = pr[it/m]) < 0 || (!used[w] && dfs(w))){
		pr[it/m] = v;
		return true;
	}
	return false;
}
int main(){
	scanf("%d%d", &n, &m);
	rep(i,n) rep(j,m){
		scanf("%d", &a);
		st[i].insert(a-1);
	}
	rep(i,m){
		fill(pr, pr+n, -1);
		rep(j,n){
			fill(used,used+n,0);
			dfs(j);
		}
		rep(j,n){
			w = pr[j];
			auto it = st[w].lower_bound(j*m);
			b[i][w] = *it;
			st[w].erase(it);
		}
	}
	rep(i,n) rep(j,m) printf("%d%c", b[j][i]+1, j<m-1?' ':'\n');
	rep(i,m) sort(b[i],b[i]+n);
	rep(i,n) rep(j,m) printf("%d%c", b[j][i]+1, j<m-1?' ':'\n');
}