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 D(...) fprintf(stderr,__VA_ARGS__)
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
const int N=105;
int n,m,a[N][N],b[N],num[N],pre[N];
bool vis[N][N],pd[N],go[N][N];
int id(int x){return (x-1)/m+1;}
bool dfs(int k1){
	if(pd[k1])return 0;
	pd[k1]=1;
	rep(i,1,n)if(go[k1][i]){
		if(!pre[i]||dfs(pre[i])){
			pre[i]=k1;
			return 1;
		}
	}
	return 0;
}
int main(){
	scanf("%d%d",&n,&m);
#ifdef xay5421
	static int p[N*N];
	rep(i,1,n*m)p[i]=i;
	mt19937 ran;
	shuffle(p+1,p+1+n*m,ran);
	rep(i,1,n)rep(j,1,m)a[i][j]=p[(i-1)*m+j];
	rep(i,1,n)rep(j,1,m){
		printf("%d%c",a[i][j],j==m?'\n':' ');
	}
#else
	rep(i,1,n)rep(j,1,m)scanf("%d",&a[i][j]);
#endif
	rep(j,1,m){
		memset(go,0,sizeof(go));
		rep(i,1,n)rep(k,j,m)go[id(a[i][k])][i]=1;
		memset(pre,0,sizeof(pre));
		rep(i,1,n){
			memset(pd,0,sizeof(pd));
			assert(dfs(i));
		}
		rep(i,1,n){
			int cur=pre[i];
			bool flg=0;
			rep(k,j,m){
				if(id(a[i][k])==cur){
					swap(a[i][k],a[i][j]);
					flg=1;
					break;
				}
			}
			assert(flg);
		}
	}
	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)b[i]=a[i][j];
		sort(b+1,b+1+n);
		rep(i,1,n)a[i][j]=b[i];
	}
	rep(i,1,n)rep(j,1,m)printf("%d%c",a[i][j],j==m?'\n':' ');
	return 0;
}
/*
5 5
4 10 22 25 2
6 20 19 13 17
5 18 14 24 12
7 3 15 1 9
16 11 21 8 23

5 4
3 10 1 15
5 19 8 12
4 9 17 13
14 18 6 7
11 20 16 2
*/