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;
typedef long long ll;
typedef pair<int,int> pp;
typedef pair<ll,ll> pll;
void cppio(){ ios_base::sync_with_stdio(0); cin.tie(0); }
#define all(x) (x).begin(),(x).end()
#define pb push_back
#define eb emplace_back
#define x first
#define y second
#define rep(i,n) for(int i = 0; i < (n); ++i)
#define rrep(i,n) for(int i = 1; i <= (n); ++i)
#define sz(x) (int)(x).size()

int n, m;

vector<int> mem[110][110];

vector<int> e[110];

int a[110][110];

int vis[110], ts;
int mat[110];
int rev[110];

bool dfs(int x) {
	if(vis[x] == ts) return false;
	vis[x] = ts;
	for(int y:e[x]) if(!rev[y] || dfs(rev[y])) {
		mat[x] = y;
		rev[y] = x;
		return true;
	}

	return false;
}

int main()
{
	cppio(); cin >> n >> m;
	rrep(i, n) {
		rrep(j, m) {
			int t; cin >> t;
			mem[i][(t+m-1)/m].pb(t);
		}
	}

	rrep(j, m) {
		rrep(i, n) {
			e[i].clear();
			e[i].reserve(m);
			rrep(k, n) {
				int c = sz(mem[i][k]);
				for(;c--;) {
					e[i].pb(k);
				}
			}
		}

		++ts;
		fill(mat+1, mat+n+1, 0);
		fill(rev+1, rev+n+1, 0);

		bool upd = 1;
		while(upd) {
			upd = 0;
			rrep(i, n) if(!mat[i] && dfs(i)) { upd = 1; }
			if(upd) ++ts;
		}

		rrep(i, n) {
			auto &v = mem[i][mat[i]];
			a[i][j] = v.back();
			v.pop_back();
		}
	}

	rrep(i, n) {
		rrep(j, m) cout << a[i][j] << ' ';
		cout << '\n';
	}

	vector<int> v;
	rrep(j, m) {
		v.clear(); rrep(i, n) v.pb(a[i][j]);
		sort(all(v));
		rrep(i, n) a[i][j] = v[i-1];
	}

	rrep(i, n) {
		rrep(j, m) cout << a[i][j] << ' ';
		cout << '\n';
	}

	return 0;
}