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 N 110

int n, m, a[N][N];

int prv[N], vis[N];
vector<int> g[N][N];

int ans[N][N];

int bpm(int x) {
	for (int i = 1; i <= n; i ++) if (!g[x][i].empty()) {
		if (vis[i]) continue;
		vis[i] = 1;
		if (!prv[i] || (prv[i] && bpm(prv[i]))) {
			prv[i] = x; return 1;
		}
	}
	return 0;
}



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]);
	for (int i = 1; i <= n; i ++) {
		for (int j = 1; j <= m; j ++) {
			int x = (a[i][j] - 1) / m + 1;
			g[i][x].push_back(a[i][j]);
		}
	}
	for (int i = 1; i <= m; i ++) {
		memset(prv, 0, sizeof prv);
		for (int j = 1; j <= n; j ++) {
			memset(vis, 0, sizeof vis);
			bpm(j);
		}
		for (int j = 1; j <= n; j ++) {
			vector<int> &cur = g[prv[j]][j];
			ans[i][prv[j]] = cur.back();
			cur.pop_back();
		}
	}
	for (int j = 1; j <= n; j ++) {
		for (int i = 1; i <= m; i ++) {
			printf("%d%c", ans[i][j], (i<m?' ':'\n'));
		}
	}
	for (int i = 1; i <= m; i ++) sort(ans[i] + 1, ans[i] + n + 1);
	for (int j = 1; j <= n; j ++) {
		for (int i = 1; i <= m; i ++) {
			printf("%d%c", ans[i][j], (i<m?' ':'\n'));
		}
	}
	return 0;
}