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:
#ifndef BZ
#pragma GCC optimize "-O3"
#endif
#include <bits/stdc++.h>

#define ALL(v) (v).begin(), (v).end()
#define rep(i, l, r) for (int i = (l); i < (r); ++i)

using ll = long long;
using ld = long double;
using ull = unsigned long long;

using namespace std;


/*
ll pw(ll a, ll b) {
	ll ans = 1; while (b) {
		while (!(b & 1)) b >>= 1, a = (a * a) % MOD;
		ans = (ans * a) % MOD, --b;
	} return ans;
}
*/

const int N = 210;

int a[N][N];

vector<pair<int, int>> eds[N];

void add_edge(int x, int y, int k) {
	eds[y].emplace_back(x, k);
}

int n, m;
int was[N];

int p[N];
int q[N];
int b[N][N];

bool dfs1(int v) {
	was[v] = 1;
	for (int i = 0; i < eds[v].size(); ++i) {
		auto e = eds[v][i];
		if (q[e.first] == -1) {
			q[e.first] = v;
			p[v] = i;
			return 1;
		}
	}
	for (int i = 0; i < eds[v].size(); ++i) {
		auto e = eds[v][i];
		if (!was[q[e.first]] && dfs1(q[e.first])) {
			q[e.first] = v;
			p[v] = i;
			return 1;
		}
	}
	return 0;
}

int main() {
	ios_base::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	cout.setf(ios::fixed), cout.precision(20);
	cin >> n >> m;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j) {
			cin >> a[i][j];
			--a[i][j];
			add_edge(a[i][j] / m, i, a[i][j]);
		}
	for (int i = 0; i < m; ++i) {
		fill(p, p + n, -1);
		fill(q, q + n, -1);
		for (int j = 0; j < n; ++j) {
			fill(was, was + n, 0);
			dfs1(j);
		}
		for (int j = 0; j < n; ++j) {
			b[i][j] = eds[j][p[j]].second;
			eds[j].erase(eds[j].begin() + p[j]);
		}
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < m; ++j)
			cout << b[j][i] + 1 << " ";
		cout << "\n";
	}
	for (int i = 0; i < m; ++i) {
		sort(b[i], b[i] + n);
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < m; ++j)
			cout << b[j][i] + 1 << " ";
		cout << "\n";
	}
	return 0;
}