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>
#define ll long long
#define ljc 998244353
using namespace std;
#define gc getchar
inline ll read(){
    register ll x=0,f=1;char ch=gc();
    while (!isdigit(ch)){if (ch=='-') f=-1;ch=gc();}
    while (isdigit(ch)){x=(x<<3)+(x<<1)+ch-'0';ch=gc();}
    return (f==1)?x:-x;
}
int n,m,col[1011010],x[1010][1010];
int vis[101010],mat[101010],matt[101010];
vector<int> e[10101]; 
int check(int u,int tim){
    for (auto v:e[u]){
        if (tim^vis[v]){
            vis[v]=tim;  
            if (!mat[v]||check(mat[v],tim)){
                mat[v]=u;matt[u]=v;return 1;
            }
        }
    }
    return 0;
}
int num[1010][1010];
void solve(int i){
	if (i>m) return;
	for (int j=1;j<=n;j++){
		e[j].clear();
		mat[j]=vis[j]=0;
		for (int k=i;k<=m;k++){
			e[j].push_back(x[j][k]);
		}
	}
	int ans=0;
	for (int j=1;j<=n;j++) ans+=check(j,j);
//	assert(ans==n);
	for (int j=1;j<=n;j++){
		if (matt[j]==x[j][i]) continue;
		for (int k=i;k<=m;k++){
			if (matt[j]==x[j][k]){
				swap(x[j][i],x[j][k]),swap(num[j][i],num[j][k]);break;
			}
		}
	}
	solve(i+1);
}
inline void mysort(int k){
	for (int i=1;i<=n;i++){
		for (int j=1;j<n;j++){
			if (num[j][k]>num[j+1][k]) swap(num[j][k],num[j+1][k]);
		}
	}
}
inline void print(){
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			printf("%d ",num[i][j]);
		}
		puts("");
	}
}
signed main(){
	n=read(),m=read();
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			num[i][j]=x[i][j]=read();
			col[(i-1)*m+j]=i;
		}
	}
	for (int i=1;i<=n;i++){
		for (int j=1;j<=m;j++){
			x[i][j]=col[x[i][j]];
		}
	}
	solve(1);
	print();
	for (int j=1;j<=m;j++) mysort(j);
	print();
	return 0;
}