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 X first
#define Y second
using namespace std;

typedef pair<int,int> pii;
const int N = 105;

int n, m, a[N][N], b[N][N], c[N][N];
int mat[N], lim, cv, vis[N];

vector<pii> adj[N];

bool match (int I) {
	if(!I) return true;
	if(vis[I] == cv) return false;
	vis[I] = cv;
	for(int i=0;i<=lim;i++) {
		int A, B;
		tie(A, B) = adj[I][i];
		if(match(mat[A])) {
			mat[A] = I;
			return true;
		}
	}
	return false;
}

int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=m;j++) {
			scanf("%d",&a[i][j]);
			adj[i].push_back({(a[i][j]-1)/m+1, a[i][j]});
		}
	}
	for(int k=1;k<=m;k++) {
		for(int i=1;i<=n;i++) {
			mat[i] = 0;
		}
		lim = m-k;
		for(int i=1;i<=n;i++) {
			++cv;
			assert(match(i));
		}
		for(int i=1;i<=n;i++) {
			for(int j=0;j<=lim;j++) {
				if(mat[adj[i][j].X] == i) {
					b[i][k] = adj[i][j].Y;
					c[adj[i][j].X][k] = adj[i][j].Y;
					swap(adj[i][j], adj[i][lim]);
					break;
				}
			}
		}
	}
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=m;j++) {
			printf("%d ", b[i][j]);
		}
		puts("");
	}
	for(int i=1;i<=n;i++) {
		for(int j=1;j<=m;j++) {
			printf("%d ", c[i][j]);
		}
		puts("");
	}
}