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;

namespace TYC
{
	typedef long long ll;
	const int N = 205;

	int n, m, s, t, ecnt, Head[N], dep[N], cur[N], R[N][N];
	struct edge { int to, next, w; } E[N * N];
	vector<int> A[N][N], P[N];

	inline void add(const int u, const int v, const int w)
	{
		E[++ecnt] = (edge){v, Head[u], w};
		Head[u] = ecnt;
		E[++ecnt] = (edge){u, Head[v], 0};
		Head[v] = ecnt;
	}	

	bool bfs()
	{
		static queue<int> Q;
		memset(dep, 0, sizeof(int[t + 1]));
		memcpy(cur, Head, sizeof(int[t + 1]));
		dep[s] = 1;
		Q.push(s);
		while (!Q.empty())
		{
			int u = Q.front();
			Q.pop();
			for (int i = Head[u], v; i; i = E[i].next)
				if (E[i].w && !dep[v = E[i].to])
					dep[v] = dep[u] + 1, Q.push(v);
		}
		return dep[t];
	}

	int dfs(const int u, const int mn)
	{
		if (u == t || !mn)
			return mn;
		int v, w, used = 0;
		for (int &i = cur[u]; i; i = E[i].next)
			if (E[i].w && dep[v = E[i].to] == dep[u] + 1)
			{
				w = dfs(v, min(E[i].w, mn - used));
				used += w;
				E[i].w -= w;
				E[i ^ 1].w += w;
				if (used == mn)
					return used;
			}
		if (!used)
			dep[u] = -1;
		return used;
	}

	void work()
	{
		scanf("%d%d", &n, &m);
		for (int i = 1, x; i <= n; i++)
			for (int j = 1; j <= m; j++)
			{
				scanf("%d", &x);
				A[i][(x - 1) / m + 1].push_back(x);
			}
		s = n << 1 | 1, t = s + 1;
		for (int k = 1; k <= m; k++)
		{
			ecnt = 1;
			memset(Head, 0, sizeof(int[t + 1]));
			for (int i = 1; i <= n; i++)
			{
				add(s, i, 1);
				add(i + n, t, 1);
				for (int j = 1; j <= n; j++)
					if (A[i][j].size())
						add(i, j + n, 1);
			}
			while (bfs())
				dfs(s, 0x3f3f3f3f);
			for (int u = 1; u <= n; u++)
				for (int i = Head[u], v; i; i = E[i].next)
					if (!E[i].w && (v = E[i].to - n) <= n)
					{
						R[u][k] = A[u][v].back();
						P[k].push_back(A[u][v].back());
						A[u][v].pop_back();
					}
			sort(P[k].begin(), P[k].end());
		}
		for (int i = 1; i <= n; i++, puts(""))
			for (int j = 1; j <= m; j++)
				printf("%d ", R[i][j]);
		for (int i = 1; i <= n; i++, puts(""))
			for (int j = 1; j <= m; j++)
				printf("%d ", P[j][i - 1]);
	}
}

int main()
{
	TYC::work();
	return 0;
}