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 ll long long
using namespace std;
int n,m,a[102][102],b[102][102],v[102],cnt[102][102],mat[102];
vector<int>zh[102][102];
int find(int x){
	int i;
	for(i=1;i<=n;i++)if(a[x][i]&&!v[i]){
		v[i]=1;
		if(!mat[i]||find(mat[i])){
			mat[i]=x;
			return 1;
		}
	}
	return 0;
}
int main(){
	int i,j,x,be,k,nt;
	scanf("%d%d",&n,&m);
	for(i=1;i<=n;i++){
		for(j=1;j<=m;j++){
			scanf("%d",&x);
			be=(x-1)/m+1;
			cnt[i][be]++;
			zh[i][be].push_back(x);
		}
	}
	for(i=1;i<=m;i++){
		//printf("%d\n",i);
		for(j=1;j<=n;j++)
		 for(k=1;k<=n;k++)a[j][k]=(cnt[j][k]!=0);
		memset(mat,0,sizeof(mat));
		for(j=1;j<=n;j++){
			memset(v,0,sizeof(v));
			assert(find(j));
		}
		//puts("-1");
		for(j=1;j<=n;j++){
			nt=mat[j];
			//printf("%d %d\n",j,nt);
			b[nt][i]=zh[nt][j].back();
			zh[nt][j].pop_back();
			cnt[nt][j]--;
		}
	}
	for(i=1;i<=n;i++){
		for(j=1;j<=m;j++)printf("%d ",b[i][j]);
		puts("");
	}
	for(i=1;i<=m;i++){
		for(j=1;j<=n;j++)
		 for(k=j+1;k<=n;k++)if(b[j][i]>b[k][i])swap(b[j][i],b[k][i]);
	}
	for(i=1;i<=n;i++){
		for(j=1;j<=m;j++)printf("%d ",b[i][j]);
		puts("");
	}
}
/*
做m次二分图匹配
不知道为什么这是对的 
*/