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;
namespace io{
	const int l=1<<20;
	char buf[l],*s,*t,c;
	inline void gc(){
		if(s==t){
			t=(s=buf)+fread(buf,1,l,stdin);
			c=s==t?EOF:*s++;
		}else c=*s++;
	}
	template<class IT>inline void gi(IT &x){
		x=0;gc();while(c<'0'||c>'9')gc();
		while('0'<=c&&c<='9'){x=(x<<1)+(x<<3)+(c^48);gc();}
	}
	char buf0[20];int a;
	template<class IT>inline void pi(IT x){
		if(x<0){putchar('-');x=-x;}
		do buf0[++a]=x%10+48;while(x/=10);
		while(a)putchar(buf0[a--]);
		putchar('\n');
	}
};
using io::gi;
using io::pi;
typedef unsigned int ui;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef pair<int,ll > pil;
typedef pair<ll ,int> pli;
typedef pair<ll ,ll > pll;
#define pque priority_queue
#define rep(i,l,r) for(i=(l);i<=(r);++i)
#define per(i,l,r) for(i=(l);i>=(r);--i)
#define REP(i,l,r) for(i=(l);i< (r);++i)
#define PER(i,l,r) for(i=(l);i> (r);--i)
#define mp make_pair
#define pb push_back
#define pf push_front
#define pob pop_back
#define pof pop_front
#define fi first
#define se second
template<class IT>inline void cmax(IT &a,IT b){if(a<b)a=b;}
template<class IT>inline void cmin(IT &a,IT b){if(b<a)a=b;}
const int N=105;
vector<int>e[N][N];
int f[N][N],a[N][N],s[N][N],p[N],vis[N],nn;
bool dfs(int v){
	int i;
	rep(i,1,nn)if(vis[i]&&f[v][i]){
		vis[i]=0;
		if(!p[i]||dfs(p[i])){p[i]=v;return true;}
	}
	return false;
}
int main(){
	int n,m,i,j,x;
	scanf("%d%d",&n,&m);nn=n;
	rep(i,1,n){
		rep(j,1,m){
			scanf("%d",&x);
			e[i][(x-1)/m+1].pb(x);
		}
	}
	rep(i,1,n)rep(x,1,n)f[i][x]=e[i][x].size();
	rep(j,1,m){
		rep(i,1,n)p[i]=0;
		rep(i,1,n){
			rep(x,1,n)vis[x]=1;
			dfs(i);
		}
		rep(i,1,n){x=p[i];s[j][x]=e[x][i][--f[x][i]];e[x][i].pob();}
	}
	rep(i,1,n){rep(j,1,m)printf("%d ",s[j][i]);printf("\n");}
	rep(j,1,m)sort(s[j]+1,s[j]+n+1);
	rep(i,1,n){rep(j,1,m)printf("%d ",s[j][i]);printf("\n");}
	return 0;
}