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 LL long long
#define SZ(x) (int)x.size()-1
#define F(i,a,b) for (int i=a;i<=b;++i)
#define DF(i,a,b) for (int i=a;i>=b;--i)
#define pb push_back
#define ms(a,b) memset(a,b,sizeof a)
using namespace std;
int read(){
    char ch=getchar(); int w=1,c=0;
    for (;!isdigit(ch);ch=getchar()) if (ch=='-') w=-1;
    for (;isdigit(ch);ch=getchar()) c=(c<<3)+(c<<1)+(ch^48);
    return w*c;
}
const int M=110;
int aim[M][M],bl[M];
vector<int> canditate[M][M];
bool used[M];
int n,m,a[M][M],b[M][M],c[M][M];
bool Fd(int x){
	F(i,1,n){
		if (canditate[x][i].size() && !used[i]){
			used[i]=1;
			if (!bl[i] || Fd(bl[i])){
				bl[i]=x;
				return 1;
			}
		}
	}
	return 0;
}
void match(int row){
	int tmp=0;
	ms(bl,0);
	for (int i=1;i<=n;i++){
		ms(used,0);
		if (Fd(i)) tmp++;
	}		
//	cout<<row<<" -------------  "<<tmp<<"\n";
	assert(tmp==n);
	F(i,1,n){
		if (bl[i]){
			int x=canditate[bl[i]][i].back();
			canditate[bl[i]][i].pop_back();
			b[bl[i]][row]=x;
			c[i][row]=x;
		}
	}
}
int main(){
	n=read(); m=read();
	F(i,1,n) F(j,1,m) {
		a[i][j]=read();
		aim[i][j]=(a[i][j]-1)/m+1;
		canditate[i][aim[i][j]].pb(a[i][j]);
	}
	F(i,1,m){
		match(i);
	}
	F(i,1,n) {
		F(j,1,m) cout<<b[i][j]<<" ";
		cout<<"\n";
	}
	F(i,1,n) {
		F(j,1,m) cout<<c[i][j]<<" ";
		cout<<"\n";
	}
    return 0;
}