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>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
#define mp make_pair
#define PI pair<int,int>
#define poly vector<ll>
#define For(i,l,r) for(int i=(int)(l);i<=(int)(r);i++)
#define Rep(i,r,l) for(int i=(int)(r);i>=(int)(l);i--)
#define pb push_back
#define fi first
#define se second
inline char gc(){
    static char buf[100000],*p1=buf,*p2=buf;
    return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
#define gc getchar
inline ll read(){
	ll x = 0; char ch = gc(); bool positive = 1;
	for (; !isdigit(ch); ch = gc())	if (ch == '-')	positive = 0;
	for (; isdigit(ch); ch = gc())	x = x * 10 + ch - '0';
	return positive ? x : -x;
}
inline void write(ll a){
    if(a<0){
    	a=-a; putchar('-');
	}
    if(a>=10)write(a/10);
    putchar('0'+a%10);
}
inline void writeln(ll a){write(a); puts("");}
inline void wri(ll a){write(a); putchar(' ');}
inline ull rnd(){
	return ((ull)rand()<<30^rand())<<4|rand()%4;
}
const int N=105;
int n,m,b[N][N],zs[N],e[N][N],match[N];
bool vis[N];
vector<int> v[N][N];
bool dfs(int p){
	For(i,1,n)if(e[p][i]&&!vis[i]){
		vis[i]=1;
		if(!match[i]||dfs(match[i])){
			match[i]=p; return 1;
		}
	}
	return 0;
}
int main(){
	n=read(),m=read();
	For(i,1,n)For(j,1,m){int x=read(); v[i][(x+m-1)/m].pb(x);}
	For(i,1,n)For(j,1,n)e[i][j]=v[i][j].size(); 
	For(j,1,m){
		memset(match,0,sizeof(match));
		For(i,1,n){memset(vis,0,sizeof(vis)); dfs(i);}
		For(i,1,n){e[match[i]][i]--; b[match[i]][j]=v[match[i]][i].back(); v[match[i]][i].pop_back();}
	}
	For(i,1,n){For(j,1,m)wri(b[i][j]); puts("");}
	For(i,1,m){For(j,1,n)zs[(b[j][i]+m-1)/m]=b[j][i]; For(j,1,n)b[j][i]=zs[j];}
	For(i,1,n){For(j,1,m)wri(b[i][j]); puts("");}
}
/*
\sum mu(d)i/j<=mid
\sum mid*i
*/