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 <cstdio>
#include <cstdlib>
#include <algorithm>
#include <cmath>
#include <vector>
#include <set>
#include <map>
#include <unordered_set>
#include <unordered_map>
#include <queue>
#include <ctime>
#include <cassert>
#include <complex>
#include <string>
#include <cstring>
#include <chrono>
#include <random>
#include <queue>
#include <bitset>
using namespace std;

#ifdef LOCAL
	#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
	#define eprintf(...) 42
#endif

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, int> pli;
typedef pair<ll, ll> pll;
typedef long double ld;
#define mp make_pair
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int N = 111;
int n, m;
int A[N][N], B[N][N], C[N][N];
vector<pii> g[N];
int mt[N];
bool used[N];

bool dfs(int x) {
	if (used[x]) return false;
	used[x] = true;
	for (pii e : g[x]) {
		int y = e.first;
		if (mt[y] == -1 || dfs(mt[y])) {
			mt[y] = x;
			return true;
		}
	}
	return false;
}
void kuhn() {
	for (int i = 0; i < n; i++)
		mt[i] = -1;
	for (int i = 0; i < n; i++)
		used[i] = false;
	for (int i = 0; i < n; i++) {
		if (dfs(i)) {
			for (int j = 0; j < n; j++)
				used[j] = false;
		}
	}
}

int main()
{
//	freopen("input.txt", "r", stdin);
//	freopen("output.txt", "w", stdout);

	scanf("%d%d", &n, &m);
	for (int i = 0; i < n; i++)
		for (int j = 0; j < m; j++) {
			scanf("%d", &A[i][j]);
			A[i][j]--;
			int x = A[i][j] / m;
			g[i].push_back(mp(x, A[i][j]));
		}
	for (int c = 0; c < m; c++) {
		kuhn();
		for (int y = 0; y < n; y++) {
			if (mt[y] == -1)
				throw;
			int x = mt[y];
			int it = 0;
			while(g[x][it].first != y) it++;
			int z = g[x][it].second;
			swap(g[x][it], g[x].back());
			g[x].pop_back();
			B[x][c] = z;
			C[y][c] = z;
		}
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++)
			printf("%d ", B[i][j] + 1);
		printf("\n");
	}
	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++)
			printf("%d ", C[i][j] + 1);
		printf("\n");
	}

	return 0;
}