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>
#define fo(i,a,b)for(int i=a,_e=b;i<=_e;++i)
using namespace std;
const int N=105;
int n,m,w,a[N][N],e[N][N];
int b[N][N],c[N][N],dy[N];
vector<int>ex[N][N];
bool us[N];
bool find(int x){
	if(us[x])return 0;
	us[x]=1;
	fo(i,1,n)if(e[x][i]&&(!dy[i]||find(dy[i]))){
		dy[i]=x;
		return 1;
	}
	return 0;
}
int main(){
	cin>>n>>m;
	fo(i,1,n)fo(j,1,m){
		scanf("%d",&a[i][j]);
		w=(a[i][j]+m-1)/m;
		e[i][w]++;
		ex[i][w].push_back(a[i][j]);
	}
	fo(i,1,m){
		fo(j,1,n)dy[j]=0;
		fo(j,1,n){
			fo(k,1,n)us[k]=0;
			find(j);
		}
		fo(j,1,n){
			c[j][i]=b[dy[j]][i]=ex[dy[j]][j].back();
			ex[dy[j]][j].pop_back();
			--e[dy[j]][j];
		}
	}
	fo(i,1,n){
		fo(j,1,m)printf("%d ",b[i][j]);
		printf("\n");
	}
	fo(i,1,n){
		fo(j,1,m)printf("%d ",c[i][j]);
		printf("\n");
	}
}