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 N = 110;
int n,m;
int a[N][N], used[N][N];
int S, T, tot, to[N * N * 2], nex[N * N * 2], final[N * 2], f[N * N * 2];
int e[N][N], ans[N][N];

void _link(int x, int y, int fl) {
	to[++tot] = y, nex[tot] = final[x], final[x] = tot;
	f[tot] = fl;
}

void link(int x, int y, int fl) {
	_link(x, y, fl), _link(y, x, 0);
}

int dis[N * 2];

void bfs() {
	for(int i = 1; i <= T; i++) dis[i] = -1;
	static int Q[N * 2]; int h = 0, t = 0;
	Q[++t] = S;
	dis[S] = 0;
	while(h < t) {
		int x = Q[++h];
		for(int i = final[x]; i; i = nex[i]) if(f[i]) {
			int y = to[i]; if (dis[y] == -1) {
				dis[y] = dis[x] + 1;
				Q[++t] = y;
			}
		}
	}
}

int go(int x,int flow) {
	if (x == T) return flow;
	int used = 0;
	for(int i = final[x]; i; i = nex[i]) if(f[i] && dis[x] + 1 == dis[to[i]]) {
		int y = to[i];
		int take = go(y, min(f[i], flow - used));
		used += take;
		f[i] -= take;
		f[i ^ 1] += take;
		if (used == flow) return used;
	}
	return used;
}

void work(int col) {
	tot = 1;
	memset(final, 0, sizeof final);
	S = n + n + 1, T = S + 1;
	for(int i = 1; i <= n; i++) {
		link(S, i, 1);
		for(int j = 1; j <= m; j++) if (!used[i][j]) {
			e[i][j] = tot + 1;
			link(i, n + 1 + (a[i][j] - 1) / m, 1);
		}
	}

	for(int i = 1; i <= n; i++) {
		link(n + i, T, 1);
	}

	while(bfs(), dis[T] != -1)
		go(S, 1e9);

	for(int i = 1; i <= n; i++) 
		for(int j = 1; j <= m; j++) if (!used[i][j]) {
			if (f[e[i][j]] == 0) {
				ans[i][col] = a[i][j];
				used[i][j] = 1;
				break;
			}
		}
}

int g[N];
int main() {
	cin>>n>>m;
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) scanf("%d", &a[i][j]);
	}

	for(int i = 1; i <= m; i++) {
		work(i);
	}

	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) printf("%d ",ans[i][j]);
		printf("\n");
	}

	for(int i = 1; i <= m; i++) {
		for(int j = 1; j <= n; j++) g[j] = ans[j][i];
		sort(g + 1, g + 1 + n);
		for(int j = 1; j <= n; j++) ans[j][i] = g[j];
	}
	
	for(int i = 1; i <= n; i++) {
		for(int j = 1; j <= m; j++) printf("%d ",ans[i][j]);
		printf("\n");
	}	
}