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 <vector>
#include <cstring>
#include <algorithm>
#define N 102

using namespace std;

int n, m;
int a[N][N], b[N][N], link[N], c[N];
bool book[N][N], vis[N], mat[N];

vector<int> e[N];

bool dfs(int);

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			scanf("%d", a[i] + j);
	for(int now = 1; now <= m; ++now){
		for(int i = 1; i <= n; ++i) e[i].clear();
		for(int i = 1; i <= n; ++i)
			for(int j = 1; j <= m; ++j){
				if(book[i][j]) continue;
				e[(a[i][j] - 1) / m + 1].push_back(i);
			}
		memset(link, 0, sizeof link);
		memset(mat, false, sizeof mat);
		for(int i = 1; i <= n; ++i){
			if(mat[i]) continue;
			memset(vis, false, sizeof vis);
			dfs(i);
		}
		for(int i = 1; i <= n; ++i){
			int p = link[i];
			for(int j = 1; j <= m; ++j){
				if(book[i][j] || (a[i][j] - 1) / m + 1 != p) continue;
				b[i][now] = a[i][j]; book[i][j] = true; break;
			}
		}
	}
	for(int i = 1; i <= n; ++i, puts(""))
		for(int j = 1; j <= m; ++j)
			printf("%d ", b[i][j]);
	for(int j = 1; j <= m; ++j){
		for(int i = 1; i <= n; ++i) c[i] = b[i][j];
		sort(c + 1, c + 1 + n);
		for(int i = 1; i <= n; ++i) b[i][j] = c[i];
	}
	for(int i = 1; i <= n; ++i, puts(""))
		for(int j = 1; j <= m; ++j)
			printf("%d ", b[i][j]);
	return 0;
}

bool dfs(int u){
	if(vis[u]) return false;;
	vis[u] = true; int sz = e[u].size();
	for(int i = 0; i < sz; ++i){
		int v = e[u][i];
		if(!link[v] || dfs(link[v])){
			mat[u] = true, link[v] = u;
			return true;
		}
	}
	return false;
}