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>
using namespace std;
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
#define CL(a,x) memset(a, x, sizeof(a))
const int N = 205;

int cap[N][N];
bool v[N];
int S,T,n,m;
int a[N>>1][N>>1],tmp[N];

int dfs(int x,int flow){
	if(v[x]) return 0;
	v[x]=true;
	if(x==T) return flow;
	int l,r;
	if(x<=n) l=n+1,r=2*n+1;
	else l=0,r=n;
	rep(y,l,r) if(cap[x][y] && !v[y]){
		int tmp=dfs(y,min(flow,cap[x][y]));
		if(tmp){
			cap[x][y]-=tmp;
			cap[y][x]+=tmp;
			return tmp;
		}
	}
	return 0;
}

int main(){
	scanf("%d%d",&n,&m);
	T = 0, S = 2 * n + 1;
	rep(i,1,n)rep(j,1,m){
		scanf("%d",a[i]+j);
		++cap[i][(a[i][j]-1)/m+n+1];
	}
	rep(k,1,m-1){
		rep(i,1,n) cap[S][i] = cap[i+n][T] = 1;
		for(CL(v,0);dfs(S,1e9);CL(v,0));
		rep(i,1,n)rep(j,n+1,2*n)if(cap[j][i]){
			cap[j][i] = 0;
			rep(x,k+1,m) if( (a[i][x]-1)/m+n+1 == j){
				swap(a[i][k], a[i][x]);
				break;
			}
		}
	}
	rep(i,1,n)rep(j,1,m)printf("%d%c",a[i][j],j==m?'\n':' ');
	rep(j,1,m){
		rep(i,1,n) tmp[i] = a[i][j];
		sort(tmp+1, tmp+n+1);
		rep(i,1,n) a[i][j] = tmp[i];
	}
	rep(i,1,n)rep(j,1,m)printf("%d%c",a[i][j],j==m?'\n':' ');
	return 0;
}