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>
using namespace std;

const int Maxn = 100;

int N, M;
vector<int> G[Maxn + 5][Maxn + 5];
int match[Maxn + 5];
bool vis[Maxn + 5], A[Maxn + 5][Maxn + 5];
int mat[Maxn + 5][Maxn + 5];

bool DFS(int u) {
	for(int v = 1; v <= N; v++)
		if(A[u][v] && !vis[v]) {
			vis[v] = true;
			if(!match[v] || DFS(match[v])) {
				match[v] = u;
				return true;
			}
		}
	return false;
}

int main() {
#ifdef LOACL
	freopen("in.txt", "r", stdin);
	freopen("out.txt", "w", stdout);
#endif
	scanf("%d %d", &N, &M);
	for(int i = 1; i <= N; i++)
		for(int j = 1; j <= M; j++) {
			int x;
			scanf("%d", &x);
			G[i][(x - 1) / M + 1].push_back(x);
		}
	for(int i = 1; i <= M; i++) {
		for(int j = 1; j <= N; j++)
			for(int k = 1; k <= N; k++)
				A[j][k] = !(G[j][k].empty());
		memset(match, 0, sizeof match);
		for(int j = 1; j <= N; j++) {
			memset(vis, false, sizeof vis);
			DFS(j);
		}
		for(int j = 1; j <= N; j++) {
			int k = match[j];
			mat[k][i] = G[k][j].back();
			G[k][j].pop_back();
		}
	}
	for(int i = 1; i <= N; i++) {
		for(int j = 1; j <= M; j++)
			printf("%d ", mat[i][j]);
		puts("");
	}
	for(int i = 1; i <= M; i++)
		for(int j = 1; j <= N; j++)
			for(int k = j + 1; k <= N; k++)
				if(mat[j][i] > mat[k][i])
					swap(mat[j][i], mat[k][i]);
	for(int i = 1; i <= N; i++) {
		for(int j = 1; j <= M; j++)
			printf("%d ", mat[i][j]);
		puts("");
	}
	return 0;
}