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 <iostream>
#include <string>
#include <cstring>
#include <vector>
#include <map>
#include <set>
#include <algorithm>
#include <math.h>
#include <cstdio>
#include <stdio.h>
#include <queue>
#include <cstdlib>
#include <deque>
#include <cassert>
#include <stack>
using namespace std;
 
#define max3(a, b, c) max(a, max(b, c))
#define min3(a, b, c) min(a, min(b, c))
#define mp make_pair
#define f first
#define se second
#define pb push_back
#define ppb pop_back
#define ll long long
#define ull unsigned long long
#define cntbit(x) __builtin_popcount(x)
#define endl '\n'
#define uset unordered_set
#define umap unordered_map
#define pii pair<int, int>
#define ld long double
#define pll pair<long long, long long>

const int inf = 2e9; 
const int N = 2e2 + 15;
int n, m, a[N][N], color[N * N], mt[N], b[N][N], cnt[N][N], c[N][N];
bool used[N];
vector <int> g[N];
stack <int> pos[N][N];

bool dfs(int v) {
	if(used[v])
		return false;
	used[v] = true;
	for(int to : g[v])
		if(cnt[v][to] && !mt[to]) {
			mt[to] = v,
			mt[v] = to;
			return true;
		}
	for(int to : g[v]) {
		if(cnt[v][to] && dfs(mt[to])) {
			mt[to] = v,
			mt[v] = to;
			return true;
		}
	}
	return false;
}

int main() {
    scanf("%d%d", &n, &m);
    for(int i = 1; i <= n; ++i)
    	for(int j = 1; j <= m; ++j) {
    		scanf("%d", &a[i][j]);
    		color[a[i][j]] = (a[i][j] - 1) / m + 1;

    		cnt[i][color[a[i][j]] + n]++;
    		cnt[color[a[i][j]] + n][i]++;

			g[i].pb(color[a[i][j]] + n);
    		g[color[a[i][j]] + n].pb(i);

    		pos[i][color[a[i][j]]].push(a[i][j]);
    	}
    for(int j = 1; j <= m; ++j) {
    	memset(mt, 0, sizeof(mt));
	    for(bool flag = 1; flag; ) {
	    	flag = false, memset(used, 0, sizeof(used));
	        for(int i = 1; i <= (n << 1); ++i)
	            if(!mt[i] && dfs(i))
	                flag = true;
	    }
    	for(int i = n + 1; i <= (n << 1); ++i)
    		if(mt[i]) {
    			cnt[mt[i]][i]--;
    			cnt[i][mt[i]]--;
    			b[mt[i]][j] = pos[mt[i]][i - n].top();
				pos[mt[i]][i - n].pop();
    		}
    }
    for(int i = 1; i <= n; ++i, puts(""))
    	for(int j = 1; j <= m; ++j) {
    		printf("%d ", b[i][j]);
    		c[color[b[i][j]]][j] = b[i][j];
    	}
    for(int i = 1; i <= n; ++i, puts(""))
    	for(int j = 1; j <= m; ++j)
    		printf("%d ", c[i][j]);
    return 0;
}