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 ll long long
#define REP(i,a,b) for(int i=(a),_end_=(b);i<=_end_;i++)
#define DREP(i,a,b) for(int i=(a),_end_=(b);i>=_end_;i--)
#define EREP(i,u) for(int i=start[u];i;i=e[i].next)
#define fi first
#define se second
#define mkr(a,b) make_pair(a,b)
#define SZ(A) ((int)A.size())
template<class T>inline void chkmin(T &a,T b){ if(a>b)a=b;}
template<class T>inline void chkmax(T &a,T b){ if(a<b)a=b;}
inline void myassert(bool a,string s){ if(!a)cerr<<s<<endl,exit(0);}
inline int read()
{
	int s=0,f=1;char ch=getchar();
	while(!isdigit(ch) && ch!='-')ch=getchar();
	if(ch=='-')ch=getchar(),f=-1;
	while(isdigit(ch))s=s*10+ch-'0',ch=getchar();
	return ~f?s:-s;
}

const int maxn=120;

int n,m,col[maxn][maxn];

vector<int>rest[maxn][maxn];

inline void init()
{
	n=read();m=read();
	REP(i,1,n)REP(j,1,m)
	{
		int x=read();
		col[i][j]=(x+m-1)/m;
		rest[i][col[i][j]].push_back(x);
	}
}

int f[maxn][maxn];
int mch[maxn],vis[maxn];

inline bool dfs(int u)
{
	REP(v,1,n)if(f[u][v] && !vis[v])
	{
		vis[v]=1;
		if(!mch[v] || dfs(mch[v]))
		{
			mch[v]=u;
			return 1;
		}
	}
	return 0;
}

int B[maxn][maxn];
int C[maxn][maxn];

inline void doing()
{
	REP(k,1,m)
	{
		REP(i,1,n)REP(j,1,n)f[i][j]=!rest[i][j].empty();
		memset(mch,0,sizeof(int)*(n+1));
		REP(i,1,n)
		{
			memset(vis,0,sizeof(int)*(n+1));
			dfs(i);
		}
		REP(i,1,n)
		{
			int x=mch[i],y=i;
			B[x][k]=rest[x][y].back(); rest[x][y].pop_back();
			C[y][k]=B[x][k];
		}
	}
	REP(i,1,n)REP(j,1,m)printf("%d%c",B[i][j],j==m?'\n':' ');
	REP(i,1,n)REP(j,1,m)printf("%d%c",C[i][j],j==m?'\n':' ');
}

int main()
{
	init();
	doing();
	return 0;
}