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;
const int maxn = 100;
int n, m;
int a[maxn + 10][maxn + 10], b[maxn + 10][maxn + 10];
bool c[maxn + 10][maxn + 10];
int lk[maxn + 10], d[maxn + 10];
bool vis[maxn + 10];
bool ban[maxn + 10][maxn + 10];

bool cmp(int x, int y) {
	return (x - 1) / m + 1 < (y - 1) / m + 1;
}

bool dfs(int p) {
	for (int i = 1; i <= n; ++i)
		if (c[p][i] && !vis[i]) {
			vis[i] = 1;
			if (!lk[i] || dfs(lk[i])) {
				lk[i] = p; 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 <= m; ++i) {
		memset(c, 0, sizeof c);
		memset(lk, 0, sizeof lk);
		for (int j = 1; j <= n; ++j)
			for (int k = 1; k <= m; ++k)
				if (!ban[j][k])
					c[j][(a[j][k] - 1) / m + 1] = 1;
		for (int j = 1; j <= n; ++j) {
			memset(vis, 0, sizeof vis);
			assert(dfs(j));
		}				
		for (int j = 1; j <= n; ++j) {
			int pos = lk[j];
			for (int k = 1; k <= m; ++k)
				if (!ban[pos][k] && (a[pos][k] - 1) / m + 1 == j) {
					ban[pos][k] = 1;
					b[pos][i] = a[pos][k];
					break;
				}
		}
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			printf("%d%c", b[i][j], " \n"[j == m]);
	for (int i = 1; i <= m; ++i) {
		for (int j = 1; j <= n; ++j)
			d[j] = b[j][i];
		sort(d + 1, d + n + 1, cmp);
		for (int j = 1; j <= n; ++j)
			b[j][i] = d[j];
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			printf("%d%c", b[i][j], " \n"[j == m]);
}