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 <iostream>
#include <cstring>
#include <cstdio>
#include <vector>
#define MN 210
#define MM 21000

int h[MN], nxt[MM], to[MM], cap[MM], K = 1;
int q[MN], hh = 0, tt = 0;
int iter[MN], d[MN];
int A[MN][MN], B[MN][MN], C[MN][MN];
std::vector<int> cnt[MN][MN];

void ins(int u, int v, int c) {nxt[++K] = h[u]; h[u] = K; to[K] = v; cap[K] = c;}
void insw(int u, int v, int c) {ins(u, v, c); ins(v, u, 0);}

bool bfs(int S, int T)
{
	memcpy(iter, h, sizeof(h));
	memset(d, 0, sizeof(d)); d[S] = 1;
	hh = tt = 0; q[tt++] = S;
	while(hh < tt)
	{
		int u = q[hh++]; 
		for(int i = h[u]; i; i = nxt[i])
			if(cap[i] && !d[to[i]])
			{
				d[to[i]] = d[u] + 1;
				q[tt++] = to[i];
			}
	}
	return d[T];
}

int dfs(int u, int T, int f)
{
	if(u == T) return f;
	int used = 0;
	for(int &i = iter[u]; i; i = nxt[i])
	{
		if(cap[i] && d[to[i]] == d[u] + 1)
		{
			int w = dfs(to[i], T, std::min(cap[i], f - used));
			if(w)
			{
				cap[i] -= w; cap[i ^ 1] += w; used += w;
				if(f == used) return f;
			}
		}
	}
	return used;
}

int dinic(int S, int T)
{
	int w, f = 0;
	while(bfs(S, T))
		while(w = dfs(S, T, 1e9)) f += w;
	return f;
}

int main()
{
	int n, m; scanf("%d%d", &n, &m);
	for(int i = 1; i <= n; i++)
		for(int j = 1; j <= m; j++)
		{
			scanf("%d", &A[i][j]);
			cnt[i][(A[i][j] - 1) / m + 1].push_back(A[i][j]);
		}
	int S = 2 * n + 1, T = 2 * n + 2;
	for(int i = 1; i <= m; i++)
	{
		memset(h, 0, sizeof(h)); K = 1;
		for(int j = 1; j <= n; j++)
			for(int k = 1; k <= n; k++)
				if(cnt[j][k].size()) insw(j, n + k, 1);
		for(int j = 1; j <= n; j++) insw(S, j, 1), insw(j + n, T, 1);
		dinic(S, T); 
		for(int j = 1; j <= n; j++)
			for(int x = h[j]; x; x = nxt[x])
				if(!cap[x] && n + 1 <= to[x] && to[x] <= 2 * n)
					B[j][i] = cnt[j][to[x] - n].back(), cnt[j][to[x] - n].pop_back();
	}
	for(int i = 1; i <= n; i++, puts(""))
		for(int j = 1; j <= m; j++)
		{
			C[(B[i][j] - 1) / m + 1][j] = B[i][j];
			printf("%d ", B[i][j]);
		}
	for(int i = 1; i <= n; i++, puts(""))
		for(int j = 1; j <= m; j++)
			printf("%d ", C[i][j]);
}