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;

#define INF int(1e9)
#define rep(i,n) for(int i=0;i<(n);i++)
#define REP(i,n) for(int i=1;i<=(n);i++)
#define mp make_pair
#define pb push_back
#define fst first
#define snd second
typedef long long ll;
typedef pair<int,int> pii;

const int maxn=105;
const int maxm=20005;
int n,m;
int a[maxn][maxn],b[maxn][maxn],c[maxn][maxn],id[maxn][maxn];
vector<int>val[maxn][maxn];

namespace Flow{
	int S,T,V,tot;
	int head[maxn*2],nxt[maxm*2],to[maxm*2],cap[maxm*2];
	int lvl[maxn*2],cur[maxn*2],q[maxn*2];
	void init(){
		tot=0;
		memset(head,-1,sizeof(head));
	}
	int addedge(int u,int v,int c){
		nxt[tot]=head[u];to[tot]=v;cap[tot]=c;head[u]=tot++;
		nxt[tot]=head[v];to[tot]=u;cap[tot]=0;head[v]=tot++;
		return tot-2;
	}
	void bfs(){
		rep(i,V)lvl[i]=-1;
		lvl[S]=0;
		int l=0,r=0;
		q[r++]=S;
		while(l<r){
			int x=q[l++];
			for(int i=head[x];~i;i=nxt[i]){
				if(cap[i]&&lvl[to[i]]<0){
					lvl[to[i]]=lvl[x]+1;
					q[r++]=to[i];
				}
			}
		}
	}
	int dfs(int x,int f){
		if(x==T)return f;
		for(int &i=cur[x];~i;i=nxt[i]){
			if(cap[i]&&lvl[to[i]]==lvl[x]+1){
				int d=dfs(to[i],min(f,cap[i]));
				if(d>0){
					cap[i]-=d;
					cap[i^1]+=d;
					return d;
				}
			}
		}
		return 0;
	}
	int work(){
		int res=0;
		for(;;){
			bfs();
			if(lvl[T]<0)break;
			rep(i,V)cur[i]=head[i];
			for(;;){
				int f=dfs(S,INF);
				if(!f)break;
				res+=f;
			}
		}
		return res;
	}
}

int main(){
	scanf("%d%d",&n,&m);
	REP(i,n)REP(j,m){
		scanf("%d",&a[i][j]);
		val[i][(a[i][j]-1)/m+1].pb(a[i][j]);
	}
	REP(i,m){
		Flow::init();
		Flow::S=0;Flow::T=n+n+1;Flow::V=Flow::T+1;
		REP(j,n)Flow::addedge(Flow::S,j,1);
		REP(j,n)Flow::addedge(j+n,Flow::T,1);
		REP(j,n)REP(k,n)if(!val[j][k].empty())id[j][k]=Flow::addedge(j,k+n,1);
		assert(Flow::work()==n);
		REP(j,n)REP(k,n)if(!val[j][k].empty()&&!Flow::cap[id[j][k]]){
			b[j][i]=val[j][k].back();
			val[j][k].pop_back();
		}
	}
	REP(i,m){
		vector<int>tmp;
		REP(j,n)tmp.pb(b[j][i]);
		sort(tmp.begin(),tmp.end());
		REP(j,n)c[j][i]=tmp[j-1];
	}
	REP(i,n)REP(j,m)printf("%d%c",b[i][j],j==m?'\n':' ');
	REP(i,n)REP(j,m)printf("%d%c",c[i][j],j==m?'\n':' ');
	return 0;
}