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 <cstdio>
#include <cstring>
#include <algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
#define fd(i,a,b) for(int i=a;i>=b;i--)
#define rep(i,a) for(int i=lst[a];i;i=nxt[i])
using namespace std;

int read() {
	char ch;
	for(ch=getchar();ch<'0'||ch>'9';ch=getchar());
	int x=ch-'0';
	for(ch=getchar();ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
	return x;
}

const int N=205,M=N*N<<1,inf=0x7fffffff;

int t[M],f[M],nxt[M],lst[N],l;

void add(int x,int y,int z) {
	t[++l]=y;f[l]=z;nxt[l]=lst[x];lst[x]=l;
	t[++l]=x;f[l]=0;nxt[l]=lst[y];lst[y]=l;
}

int n,m,a[N][N],b[N][N],c[N][N];
bool used[N][N];

int P(int x) {return (x-1)/m+1;}

int S,T,dis[N],q[N];

bool bfs() {
	fo(i,S,T) dis[i]=-1;dis[S]=0;
	int i=0,j=1;q[1]=S;
	while (i<j) 
		rep(k,q[++i])
			if (f[k]&&dis[t[k]]==-1) {
				dis[t[k]]=dis[q[i]]+1;
				q[++j]=t[k];
			}
	return dis[T]!=-1;
}

int dinic(int x,int y) {
	if (x==T) return y;
	int now=0;
	rep(i,x)
		if (f[i]&&dis[t[i]]==dis[x]+1) {
			int k=dinic(t[i],min(y,f[i]));
			y-=k;now+=k;f[i]-=k;f[i^1]+=k;
			if (!y) break;
		}
	if (!now) dis[x]=-1;
	return now;
}

int main() {
	n=read();m=read();
	fo(i,1,n) fo(j,1,m) a[i][j]=read();
	S=0;T=n<<1|1;
	fo(w,1,m) {
		l=1;fo(i,S,T) lst[i]=0;
		fo(i,1,n)
			fo(j,1,m)
				if (!used[i][j]) 
					add(i,n+P(a[i][j]),1);
		fo(i,1,n) add(S,i,1);
		fo(i,1,n) add(i+n,T,1);
		while (bfs()) dinic(S,inf);
		int p=2;
		fo(i,1,n)
			fo(j,1,m)
				if (!used[i][j]) {
					if (!f[p]) {
						used[i][j]=1;
						b[i][w]=a[i][j];
					}
					p+=2;
				}
	}
	fo(i,1,n) {
		fo(j,1,m) printf("%d ",b[i][j]);
		puts("");
	}
	fo(j,1,m) fo(i,1,n) c[P(b[i][j])][j]=b[i][j];
	fo(i,1,n) {
		fo(j,1,m) printf("%d ",c[i][j]);
		puts("");
	}
	return 0;
}