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 <vector>
#include <string>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <random>
#include <chrono>
#include <tuple>
#include <random>
#include <cmath>

using namespace std;

typedef long long ll;
typedef long double ld;
#define fastInp cin.tie(0); cout.tie(0); ios_base::sync_with_stdio(0);

const ll SZ = 1009, SZ2 = 1e5;

ll grid[SZ][SZ], us[SZ][SZ];
pair<ll, ll> gr[SZ][SZ], ans[SZ][SZ];

ll mt[SZ2], used[SZ2];
ll cur = 0;

vector<vector<ll>> g;

bool khun(int v) {
	used[v] = cur;

	for (auto to : g[v]) {
		if (mt[to] == -1) {
			mt[to] = v;
			return true;
		}
		else if (used[mt[to]] != cur && khun(mt[to])) {
			mt[to] = v;
			return true;
		}
	}

	return false;
}

int main()
{
	fastInp;

	ll n, m;
	cin >> n >> m;

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			cin >> grid[i][j];
			grid[i][j]--;
			gr[i][j] = { grid[i][j] / m, grid[i][j] % m };
		}
	}

	for (int it = 0; it < m; it++) {
		g.clear();
		for (int j = 0; j < m + n; j++) {
			mt[j] = -1;
		}
		for (int i = 0; i < n + m; i++) {
			used[i] = 0;
		}
		cur = 1;

		ll cnt = 0;
		vector<ll> rws;
		vector<pair<ll, ll>> crds;
		g.resize(n);
		map<pair<ll, ll>, pair<ll, ll>> c;
		for (int i = 0; i < n; i++) {
			for (int j = 0; j < m; j++) {
				if (!us[i][j]) {
					g[i].push_back(gr[i][j].first);
					c[{gr[i][j].first, i}] = { i, j };
				}
			}
		}

		for (int i = 0; i < n; i++) {
			khun(i);
			cur++;
		}

		for (int j = 0; j < n; j++) {
			ll vl = mt[j];
			pair<ll, ll> crds = c[{j, vl}];
			us[crds.first][crds.second] = 1;
			ans[vl][it] = gr[crds.first][crds.second];
		}
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			cout << ans[i][j].first * m + ans[i][j].second + 1 << " ";
		}
		cout << "\n";
	}

	for (int i = 0; i < n; i++) {
		for (int j = 0; j < m; j++) {
			for (int k = i + 1; k < n; k++) {
				if (ans[i][j].first > ans[k][j].first) swap(ans[i][j], ans[k][j]);
			}
			cout << ans[i][j].first * m + ans[i][j].second + 1 << " ";
		}
		cout << "\n";
	}

	return 0;
}