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:
//khodaya khodet komak kon
#include <bits/stdc++.h>

#define F first
#define S second
#define pb push_back
#define all(x) x.begin(), x.end()
#pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")


using namespace std;

typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int N = 100 + 10;
const ll MOD = 1000000000 + 7;
const ll INF = 1000000010;
const ll LOG = 25;

int a[N][N], n, m, M[N];
int mark[N], MT[N];
vi G[N];

bool DFS(int v){
	mark[v] = 1;
	for (auto u:G[v]){
		if (mark[M[u]]) continue;
		if (M[u] == 0){
			M[u] = v;
			MT[v] = 1;
			return 1;
		}
		if (DFS(M[u])){
			M[u] = v;
			MT[v] = 1;
			return 1;
		}
	}
	return 0;
}

inline void Matching(){
	memset(M, 0, sizeof M);
	memset(MT, 0, sizeof MT);
	for (int i = 1; i <= n; i++){
		memset(mark, 0, sizeof mark);
		for (int j = 1; j <= n; j++){
			if (!mark[j] && !MT[j]){
				if (DFS(j)){
					break;
				}
			}
		}
	}
//	for (int i = 1; i <= n; i++) cout << M[i] << ' ';
//	cout << '\n';
}

inline void Solve(int x){
	for (int i = 1; i <= n; i++) G[i].clear();
//	cout << "Fuck\n";
	for( int i = 1; i <= n; i++) for (int j = x; j <= m; j++){
		G[a[i][j] / m + 1].pb(i);
//		cout << a[i][j] / m + 1 << ' ' << i << '\n';
	}
//	cout << "YES" << endl;
	Matching();
//	cout << "YES2" << endl;
	for (int i = 1; i <= n; i++){
		for (int j = x; j <= m; j++){
			if (a[i][j] / m + 1 == M[i]){
				swap(a[i][x], a[i][j]);
				break;
			}
		}
	}
}

int32_t main(){
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	cin >> n >> m;
	for (int i = 1; i <= n; i++) for (int j = 1; j <= m; j++){
		cin >> a[i][j];
		a[i][j] --;
	}
	for (int i = 1; i <= m; i++){
		Solve(i);		
	}
	for (int i = 1; i <= n; i++){
		for (int j = 1; j <= m; j++) cout << a[i][j] + 1 << ' ';
		cout << '\n';
	}
	for (int i = 1; i <= m; i++){
		for (int j = 1; j <= n; j++){
			for (int k = j; k <= n; k++){
				if (a[k][i] / m + 1 == j){
					swap(a[j][i], a[k][i]);
					break;
				}
			}
		}
	}
	for (int i = 1; i <= n; i++){
		for (int j = 1; j <= m; j++) cout << a[i][j] + 1 << ' ';
		cout << '\n';
	}







	return 0;
}