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<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<vector>
using namespace std;
#define MAXN 400
#define MAXM 160000
#define LL long long
#define Pr pair<int,int>
#define X first
#define Y second
LL read(){
    LL x=0,F=1;char c=getchar();
    while(c<'0'||c>'9'){if(c=='-')F=-1;c=getchar();}
    while(c>='0'&&c<='9'){x=x*10+c-'0';c=getchar();}
    return x*F;
}
int n,m,tot;
int a[MAXM+5],vis[MAXM+5],tmp[MAXM+5];
int match[MAXM+5],ans[MAXN+5][MAXN+5],tp[MAXN+5][MAXN+5];
vector<int> P[MAXN+5][MAXN+5];
vector<int> G[MAXM+5];
int dfs(int x)
{
	if(vis[x])return 0;
	vis[x]=1;
    for(int i=0;i<G[x].size();i++){
        int xnt=G[x][i];
        if(match[xnt]==-1||dfs(match[xnt])){
            match[xnt]=x;
            return 1;
        }
    }return 0;
}
int main()
{
	//freopen("sort.in","r",stdin);
	//freopen("sort.out","w",stdout);
	n=read(),m=read();
	for(int i=1;i<=n;i++)
		for(int j=1;j<=m;j++){
			int x=read();
			a[x]=i;//在第i行
			P[i][(x-1)/m+1].push_back(x);
		}
	for(int t=1;t<=m;t++){//m次最大匹配O(n^4) 
		for(int i=1;i<=n;i++)
		G[i].clear(),match[i]=-1;
		for(int i=1;i<=n;i++){
			for(int j=1;j<=n;j++)
			if(!P[i][j].empty())G[i].push_back(j);
		}
		for(int i=1;i<=n;i++){
			memset(vis,0,sizeof(vis));
			dfs(i);
		}
		for(int i=1;i<=n;i++){
			ans[match[i]][t]=P[match[i]][i].back();
			P[match[i]][i].pop_back();
		}
	}
	for(int i=1;i<=n;i++,puts(""))
		for(int j=1;j<=m;j++)
			printf("%d ",ans[i][j]);
	for(int i=1;i<=m;i++){
		for(int j=1;j<=n;j++)
		tmp[j]=ans[j][i];
		sort(tmp+1,tmp+n+1);
		for(int j=1;j<=n;j++)
		tp[j][i]=tmp[j];
	}
	for(int i=1;i<=n;i++,puts(""))
		for(int j=1;j<=m;j++)
			printf("%d ",tp[i][j]);
}