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 rep(i,a,b) for(int i=(a);i<=(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define REP(i,n) for(int i=(0);i<(n);i++)
#define fi first
#define se second
#define pb push_back
#define mp make_pair
using namespace std;
typedef pair<int,int> pii;
typedef vector<int> vi;
typedef long long ll;

template<class T> inline void read(T &x){
	int f=0;x=0;char ch=getchar();
	for(;!isdigit(ch);ch=getchar())f|=(ch=='-');
	for(;isdigit(ch);ch=getchar())x=x*10+ch-'0';
	if(f)x=-x;
}

const int N=105;
namespace G{
	int f[N][N],x[N],y[N],vis[N],n,tim;
	inline void init(int _n){
		n=_n,tim=1;
		memset(f,0,sizeof f);
		memset(vis,0,sizeof vis);
		memset(x,0,sizeof x);
		memset(y,0,sizeof y);
	}
	int dfs(int u){
		vis[u]=tim;
		rep(v,1,n)
			if(f[u][v]){
				if(!y[v]||vis[y[v]]!=tim&&dfs(y[v])){
					y[v]=u,x[u]=v;
					return 1;
				}
			}
		return 0;
	}
	inline void solve(){
		rep(i,1,n)
			if(!x[i]){
				dfs(i);
				tim++;
			}
		// cerr<<tim<<endl;
	}
}
int bl[N][N],a[N][N],b[N][N],c[N][N],used[N][N];
int n,m;


int main(){
	read(n),read(m);
	rep(i,1,n)
		rep(j,1,m){
			read(a[i][j]);
			bl[i][j]=(a[i][j]-1)/m+1;
		}
	rep(j,1,m){
		G::init(n);
		rep(i,1,n)rep(k,1,m)
			G::f[i][bl[i][k]]|=!used[i][k];
		G::solve();
		rep(i,1,n){
			int cur=0;
			rep(k,1,m)
				if(!used[i][k]&&G::x[i]==bl[i][k]){
					cur=k;break;
				}
			used[i][cur]=1;
			b[i][j]=a[i][cur];
			c[bl[i][cur]][j]=a[i][cur];
		}
	}
	rep(i,1,n){
		rep(j,1,m)
			printf("%d ",b[i][j]);
		puts("");
	}
	rep(i,1,n){
		rep(j,1,m)
			printf("%d ",c[i][j]);
		puts("");
	}
	return 0;
}