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:
// zhe fa neng guo
#include <bits/stdc++.h>
using namespace std;

const int maxn = 100;
int n, m, a[maxn + 3][maxn + 3], c[maxn + 3][maxn + 3], tmp[maxn + 3];
int row[maxn + 3], col[maxn + 3], ans[maxn + 3][maxn + 3];
bool used[maxn + 3][maxn + 3], vis[maxn + 3];

bool match(int u) {
	for (int i = 1, v; i <= m; i++) {
		if (!used[u][i]) {
			v = c[u][i];
			if (vis[v]) {
				continue;
			}
			vis[v] = true;
			if (!row[v] || match(row[v])) {
				row[v] = u, col[v] = 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]);
			c[i][j] = (a[i][j] - 1) / m + 1;
		}
	}
	for (int i = 1; i <= m; i++) {
		memset(row, 0, sizeof(row));
		memset(col, 0, sizeof(col));
		for (int j = 1; j <= n; j++) {
			memset(vis, false, sizeof(vis));
			assert(match(j));
		}
		for (int j = 1; j <= n; j++) {
			ans[row[j]][i] = a[row[j]][col[j]];
			used[row[j]][col[j]] = true;
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			printf("%d%c", ans[i][j], " \n"[j == m]);
		}
	}
	for (int i = 1; i <= m; i++) {
		for (int j = 1; j <= n; j++) {
			tmp[j] = ans[j][i];
		}
		sort(tmp + 1, tmp + n + 1);
		for (int j = 1; j <= n; j++) {
			ans[j][i] = tmp[j];
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			printf("%d%c", ans[i][j], " \n"[j == m]);
		}
	}
	return 0;
}