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 <cstdio>
#include <cstring>
#include <vector>
#include <set>
#include <cstdlib>
#include <map>
#include <iostream>
#include <algorithm>

#define X first
#define Y second
#define PB push_back

using namespace std;

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

const int N = 105;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const int BASE = 31337;

int con[N][N], bio[N], mat_L[N], mat_R[N], n, m, A[N][N], B[N][N];
vector < int > trpam[N][N];

bool dfs(int x){
	if(bio[x]) return 0;
	bio[x] = 1;
	for(int i = 0;i < n;i++){
		if(!con[x][i]) continue;
		if(mat_R[i] == -1 || dfs(mat_R[i])){
			mat_L[x] = i; mat_R[i] = x;
			return 1;
		}
	}
	return 0;
} 

int main(){
	scanf("%d%d", &n, &m);
	for(int i = 0;i < n;i++){
		for(int j = 0;j < m;j++){
			scanf("%d", &A[i][j]);
			A[i][j]--;
			con[i][A[i][j] / m]++;
			trpam[i][A[i][j] / m].PB(A[i][j]);
		}
	}
	for(int j = 0;j < m;j++){
		memset(mat_L, -1, sizeof(mat_L));
		memset(mat_R, -1, sizeof(mat_R));
		int cnt = 0;
		for(int i = 0;i < n;i++){
			memset(bio, 0, sizeof(bio));
			cnt += dfs(i);
		}
		if(cnt != n)
			printf("kriticno\n");
		for(int i = 0;i < n;i++){
			B[i][j] = trpam[i][mat_L[i]].back();
			con[i][mat_L[i]]--;
			trpam[i][mat_L[i]].pop_back();
		}
	}
	for(int i = 0;i < n;i++){
		for(int j = 0;j < m;j++){
			printf("%d ", B[i][j] + 1);
		}
		printf("\n");
	}
	for(int j = 0;j < m;j++){
		vector < int > tmp;
		for(int i = 0;i < n;i++)
			tmp.PB(B[i][j]);
		sort(tmp.begin(), tmp.end());
		for(int i = 0;i < n;i++)
			B[i][j] = tmp[i];
	}
	for(int i = 0;i < n;i++){
		for(int j = 0;j < m;j++){
			printf("%d ", B[i][j] + 1);
		}
		printf("\n");
	}
	return 0;
}