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>

const int N = 105;
int n, m, A[N][N], cnt[N], val[N], p[N], bel[N], ibel[N];
bool edge[N][N], vis[N];
bool find(int x) {
	for (int i = 0; i < n; ++i) if (!vis[i] && edge[x][i]) {
		vis[i] = 1;
		if (!~bel[i] || find(bel[i])) {
			bel[i] = x; return 1;
		}
	}
	return 0;
}
void solve(int col) {
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < n; ++j) edge[i][j] = 0;
	for (int i = 0; i < n; ++i) {
		for (int j = col; j < m; ++j)
			edge[i][A[i][j] / m] = 1;
		bel[i] = -1;
	}
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) vis[j] = 0;
		assert(find(i));
	}
	for (int i = 0; i < n; ++i) ibel[bel[i]] = i;
	for (int i = 0; i < n; ++i)
		for (int j = col; j < m; ++j)
			if (A[i][j] / m == ibel[i])
				std::swap(A[i][j], A[i][col]);
}
int main() {
	std::ios::sync_with_stdio(0), std::cin.tie(0);
	std::cin >> n >> m;
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j) std::cin >> A[i][j], --A[i][j];
	for (int i = 0; i < m; ++i) solve(i);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j)
			std::cout << A[i][j] + 1 << " \n"[j == m - 1];
	for (int i = 0; i < m; ++i)
		for (int j = 0; j < n; ++j)
			for (int k = j + 1; k < n; ++k)
				if (A[j][i] > A[k][i]) std::swap(A[j][i], A[k][i]);
	for (int i = 0; i < n; ++i)
		for (int j = 0; j < m; ++j)
			std::cout << A[i][j] + 1 << " \n"[j == m - 1];
	return 0;
}