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:
#pragma warning (disable:4996)
#include"bits/stdc++.h"
#include<cassert>
#define int long long
#define MRE assert(0);
#define rep(i,n) for(int i=0;i<n;i++)
using namespace std;
const long long mod = 1000000007;
const long long inf = 1e17;
typedef pair<int, int> P;
struct edge {
	int to, id;
};
int a[105][105];
int match[205]; bool used[205];
vector<int>ans[205],ans2[205];
vector<int>G1[105][105];
vector<int>G[205];

void add_edge(int u, int v) {
	G[u].push_back(v);
	G[v].push_back(u);
}

bool dfs(int v) {
	used[v] = true;
	for (int i : G[v]) {
		int u = i, w = match[u];
		if (w < 0 || !used[w] && dfs(w)) {
			match[v] = u; match[u] = v; return true;
		}
	}
	return false;
}

void matching(int n) {
	rep(i, n)G[i].clear();
	rep(i, n) {
		rep(j, n) {
			if (G1[i][j].size())add_edge(i, n+j);
		}
	}
	memset(match, -1, sizeof(match));
	rep(i, n) {
		if (match[i] < 0) {
			memset(used, 0, sizeof(used));
			dfs(i);
		}
	}
	rep(i, n) {
		ans[i].push_back(G1[i][match[i] - n][0]);
		G1[i][match[i] - n].erase(G1[i][match[i] - n].begin());
	}
}

signed main() {
	int n, m; cin >> n >> m;

	rep(i, n) {
		rep(j, m) {
			cin >> a[i][j];
			G1[i][(a[i][j]-1)/m].push_back(a[i][j]);
		}
	}

	rep(i,m)matching(n);
	rep(i, n) {
		rep(j, m) {
			if (j)cout << ' '; cout << ans[i][j];
		}cout << endl;
	}
	rep(j, m) {
		vector<int>s;
		rep(i, n)s.push_back(ans[i][j]);
		sort(s.begin(), s.end());
		rep(i, n)ans2[i].push_back(s[i]);
	}
	rep(i, n) {
		rep(j, m) {
			if (j)cout << ' '; cout << ans2[i][j];
		}cout << endl;
	}
}