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 clr(x) memset(x,0,sizeof x)
#define For(i,a,b) for (int i=(a);i<=(b);i++)
#define Fod(i,b,a) for (int i=(b);i>=(a);i--)
#define pb(x) push_back(x)
#define mp(x,y) make_pair(x,y)
#define fi first
#define se second
#define outval(x) cerr<<#x" = "<<x<<endl
#define outtag(x) cerr<<"-----------------"#x"-----------------\n"
#define outarr(a,L,R) cerr<<#a"["<<L<<".."<<R<<"] = ";\
                    For(_x,L,R) cerr<<a[_x]<<" ";cerr<<endl;
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair <int,int> pii;
LL read(){
    LL x=0,f=0;
    char ch=getchar();
    while (!isdigit(ch))
        f=ch=='-',ch=getchar();
    while (isdigit(ch))
        x=(x<<1)+(x<<3)+(ch^48),ch=getchar();
    return f?-x:x;
}
const int N=105;
int n,m;
int a[N][N],b[N];
int g[N][N];
int vis[N],Match[N];
int dfs(int x){
	For(y,1,n)
		if (g[x][y]&&!vis[y]){
			vis[y]=1;
			if (!Match[y]||dfs(Match[y])){
				Match[y]=x;
				return 1;
			}
		}
	return 0;
}
int main(){
	n=read(),m=read();
	For(i,1,n)
		For(j,1,m)
			a[i][j]=read();
	For(c,1,m){
		clr(g);
		For(i,1,n)
			For(j,c,m){
				int r=(a[i][j]+m-1)/m;
				if (!g[i][r])
					g[i][r]=j;
			}
		clr(Match);
		For(i,1,n){
			clr(vis);
			assert(dfs(i));
		}
		For(i,1,n){
			int r=Match[i];
			swap(a[r][c],a[r][g[r][i]]);
		}
	}
	For(i,1,n){
		For(j,1,m)
			printf("%d ",a[i][j]);
		puts("");
	}
	For(i,1,m){
		For(j,1,n)
			b[j]=a[j][i];
		sort(b+1,b+n+1);
		For(j,1,n)
			a[j][i]=b[j];
	}
	For(i,1,n){
		For(j,1,m)
			printf("%d ",a[i][j]);
		puts("");
	}
	return 0;
}