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;
#define rep(i,a,b) for(int i = (a); i < (b); ++i)
#define rrep(i,a,b) for(int i = (b); i --> (a);)
#define all(v) (v).begin(),(v).end()
#define trav(x,v) for(auto &x : v)
#define sz(v) int(v.size())
typedef vector<int> vi;
typedef long long ll;
typedef pair<int,int> pii;

bool find(int j, vector<vi>& g, vi& btoa, vi& vis) {
	if (btoa[j] == -1) return 1;
	vis[j] = 1; int di = btoa[j];
	trav(e, g[di])
		if (!vis[e] && find(e, g, btoa, vis)) {
			btoa[e] = di;
			return 1;
		}
	return 0;
}
int dfsMatching(vector<vi>& g, vi& btoa) {
	vi vis;
	rep(i,0,sz(g)) {
		vis.assign(sz(btoa), 0);
		trav(j,g[i])
			if (find(j, g, btoa, vis)) {
				btoa[j] = i;
				break;
			}
	}
	return sz(btoa) - (int)count(all(btoa), -1);
}

int n, m;

vector<vi> step1(vector<vi> &as){
	vector<vi> bs(n, vi(m));
	vector<vector<vi>> lists(n, vector<vi>(n));
	rep(i,0,n) rep(j,0,m)
		lists[i][ (as[i][j]-1)/m ].push_back(j);
	rep(c,0,m){
		vector<vi> g(n);
		rep(l,0,n) rep(r,0,n)
			if(!lists[r][l].empty())
				g[l].push_back(r);
		vi btoa(n, -1);
		dfsMatching(g, btoa);
		rep(r,0,n){
			int l = btoa[r];
			int j = lists[r][l].back();
			lists[r][l].pop_back();
			bs[r][c] = as[r][j];
		}
	}
	return bs;
}

vector<vi> step2(vector<vi> &bs){
	vector<vi> cs(n, vi(m));
	rep(i,0,n) rep(j,0,m){
		int r = (bs[i][j]-1) / m;
		cs[r][j] = bs[i][j];
	}
	return cs;
}

void skriv(vector<vi> &xs){
	trav(v, xs){
		trav(x, v) cout << x << ' ';
		cout << endl;
	}
}

int main(){
	cin.sync_with_stdio(0);
	cin.tie(0);

	cin >> n >> m;

	vector<vi> as(n, vi(m));
	trav(v, as) trav(x, v) cin >> x;

	vector<vi> bs = step1(as);
	vector<vi> cs = step2(bs);
	skriv(bs);
	puts("");
	skriv(cs);
}