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 <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
using namespace std;
typedef long long LL;

vector <LL> v[405][405];
LL n,m;
LL vis[405],p[405];
LL e[405][405],ans[405][405],a[405][405];

LL dfs(LL u){
	if(!u) return 1;
	for(LL i = 1;i <= n + n;i ++){
		if(e[u][i] <= 0) continue;
		if(vis[i]) continue;
		vis[i] = 1;
		if(dfs(p[i])){
			p[u] = i; p[i] = u;
			return 1;
		}
	}
	return 0;
}

void get_match(LL x){
	memset(p,0,sizeof(p));
	for(LL i = 1;i <= n;i ++){
		memset(vis,0,sizeof(vis));
		dfs(i);
	}
	for(LL i = 1;i <= n;i ++){
		e[i][p[i]] --; e[p[i]][i] --;
		ans[x][i] = v[i][p[i]][v[i][p[i]].size() - 1];
		v[i][p[i]].pop_back();
	}
}

int main(){
	ios::sync_with_stdio(false);
	cin >> n >> m;
	for(LL i = 1;i <= n;i ++){
		for(LL j = 1;j <= m;j ++){
			cin >> a[i][j];
			e[i][(a[i][j] - 1) / m + 1 + n] ++;
			e[(a[i][j] - 1) / m + 1 + n][i] ++;
			v[i][(a[i][j] - 1) / m + 1 + n].push_back(a[i][j]);
		}
	}
	for(LL i = 1;i <= m;i ++) get_match(i);
	for(LL i = 1;i <= n;i ++){
		for(LL j = 1;j <= m;j ++){
			cout << ans[j][i] << (j == m ? '\n' : ' ');
		}
	}
	for(LL i = 1;i <= m;i ++){
		for(LL j = 1;j <= n;j ++){
			for(LL k = 1;k <= n;k ++){
				if(ans[i][j] < ans[i][k]) swap(ans[i][j],ans[i][k]);
			}
		}
	}
	for(LL i = 1;i <= n;i ++){
		for(LL j = 1;j <= m;j ++){
			cout << ans[j][i] << (j == m ? '\n' : ' ');
		}
	}
	return 0;
}