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 <cassert>
#include <cstring>
#include <algorithm>
#define meow(args...) fprintf(stderr, args)
bool vis[100];
int n, m, a[100][100], e[100][100], ans[100][100], back[100];
std::vector<int> num[100][100];
bool match(int u) {
	for(int v=0; v<n; ++v)
		if(e[u][v]&&!vis[v]) {
			vis[v]=true;
			if(back[v]==-1||match(back[v])) {
				back[v]=u;
				return true;
			}
		}
	return false;
}
int main() {
	assert(scanf("%d%d", &n, &m)==2);
	for(int i=0; i<n; ++i)
		for(int j=0; j<m; ++j) {
			assert(scanf("%d", a[i]+j)==1);
			++e[i][(a[i][j]-1)/m];
			num[i][(a[i][j]-1)/m].push_back(a[i][j]);
		}
	for(int i=0; i<m; ++i) {
		memset(back, -1, n*sizeof(int));
		for(int j=0; j<n; ++j) {
			memset(vis, 0, n);
			assert(match(j));
		}
		for(int j=0; j<n; ++j) {
			int k=back[j];
			assert(~k);
			ans[i][k]=num[k][j][--e[k][j]];
		}
	}
	for(int i=0; i<n; ++i) {
		for(int j=0; j<m; ++j) printf("%d ", ans[j][i]);
		puts("");
	}
	for(int i=0; i<m; ++i) std::sort(ans[i], ans[i]+n);
	for(int i=0; i<n; ++i) {
		for(int j=0; j<m; ++j) printf("%d ", ans[j][i]);
		puts("");
	}
	return 0;
}