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 Maxn = 305;
int n, m, s, t, cnt, cur[Maxn], col[Maxn * Maxn], dis[Maxn], tmp[Maxn], head[Maxn], a[Maxn][Maxn], ans1[Maxn][Maxn], ans2[Maxn][Maxn];
vector <int> Ve[Maxn][Maxn];
struct edg
{
	int nxt, to, w;
	bool del;
}edge[2 * Maxn * Maxn];
void add(int x, int y, int w)
{
	edge[++cnt] = (edg){head[x], y, w};
	head[x] = cnt;
	edge[++cnt] = (edg){head[y], x, 0, true};
	head[y] = cnt;
}
queue <int> Qu;
bool bfs(void)
{
	memset(dis, 0, sizeof(dis));
	Qu.push(s);
	while (!Qu.empty())
	{
		int u = Qu.front();
		Qu.pop();
		for (int i = head[u]; i; i = edge[i].nxt)
		{
			int to = edge[i].to;
			if (edge[i].w && to != s && !dis[to])
				dis[to] = dis[u] + 1, Qu.push(to);
		}
	}
	return (bool)dis[t];
}
int dfs(int u, int mini)
{
	if (u == t || !mini) return mini;
	int w, used = 0;
	for (int &i = cur[u]; i; i = edge[i].nxt)
	{
		int to = edge[i].to;
		if (dis[to] == dis[u] + 1 && edge[i].w)
		{
			w = dfs(to, min(edge[i].w, mini - used));
			edge[i].w -= w, edge[((i - 1) ^ 1) + 1].w += w, used += w;
		}
	}
	return used;
}
void dinic(void)
{
	while (bfs())
	{
		memcpy(cur, head, sizeof(cur));
		dfs(s, 0x3f3f3f3f);
	}
}
int main()
{
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n * m; i++)
		col[i] = (i - 1) / m + 1;
	for (int i = 1; i <= n; i++)
		for (int j = 1; j <= m; j++)
		{
			scanf("%d", &a[i][j]);
			add(i, n + col[a[i][j]], 1);
			Ve[i][col[a[i][j]]].push_back(a[i][j]);
		}
	s = 2 * n + 1, t = 2 * n + 2;
	for (int i = 1; i <= m; i++)
	{
		for (int j = 1; j <= n; j++)
			add(s, j, 1), add(n + j, t, 1);
		dinic();
		for (int u = 1; u <= n; u++)
			for (int j = head[u]; j; j = edge[j].nxt)
			{
				int to = edge[j].to;
				if (!edge[j].w && !edge[j].del)
				{
					edge[j].del = true;
					edge[((j - 1) ^ 1) + 1].w = 0;
					ans1[u][i] = *Ve[u][to - n].rbegin();
					Ve[u][to - n].pop_back();
					break;
				}
			}
	}
	for (int i = 1; i <= m; i++)
	{
		for (int j = 1; j <= n; j++)
			tmp[j] = ans1[j][i];
		sort(tmp + 1, tmp + 1 + n);
		for (int j = 1; j <= n; j++)
			ans2[j][i] = tmp[j];
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
			printf("%d ", ans1[i][j]);
		puts("");
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= m; j++)
			printf("%d ", ans2[i][j]);
		puts("");
	}
	return 0;
}