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 For(i,j,k) for (int i=(int)(j);i<=(int)(k);i++)
#define Rep(i,j,k) for (int i=(int)(j);i>=(int)(k);i--)
#define pii pair<int,int>
#define pll pair<ll,ll>
#define ll long long
#define fi first
#define se second
#define PB push_back
#define uint unsigned
#define ull unsigned ll 
using namespace std;
const int N=105;
vector<int> vec[N][N];
int cnt[N][N],n,m;
int e[N][N],a[N][N];
int mat[N],vis[N];
bool find(int x){
	For(i,1,n)
		if (e[x][i]&&!vis[i]){
			vis[i]=1;
			if (!mat[i]||find(mat[i])){
				mat[i]=x;
				return 1;
			}
		}
	return 0;
} 
int main(){
	scanf("%d%d",&n,&m);
	For(i,1,n) For(j,1,m){
		int x;
		scanf("%d",&x);
		int bel=(x-1)/m+1;
		++cnt[i][bel];
		vec[i][bel].PB(x);
	}
	For(i,1,m){
		For(j,1,n) For(k,1,n){
			e[j][k]=(cnt[j][k]!=0);
		}
		memset(mat,0,sizeof(mat));
		For(j,1,n){
			memset(vis,0,sizeof(vis));
			assert(find(j));
		}
		For(j,1,n){
			int to=mat[j];
			a[to][i]=vec[to][j].back();
			vec[to][j].pop_back();
			--cnt[to][j];
		}
	}
	For(i,1,n){
		For(j,1,m) printf("%d ",a[i][j]);
		puts("");
	}
	For(i,1,m){
		For(j,1,n) For(k,j+1,n)
			if (a[j][i]>a[k][i])
				swap(a[j][i],a[k][i]);
	}
	For(i,1,n){
		For(j,1,m) printf("%d ",a[i][j]);
		puts("");
	}
}