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;

int n, m, A[105][105], tot, nxt[200 * 105], fst[205], cur[205], to[200 * 105], f[200 * 105], h[205];

void addedge(int u, int v, int w) {
	nxt[++tot] = fst[u]; fst[u] = tot; to[tot] = v; f[tot] = w;
	nxt[++tot] = fst[v]; fst[v] = tot; to[tot] = u; f[tot] = 0;
}
bool bfs() {
	static int q[205];
	int st = 0, ed = 0;
	for (int i = 1; i <= 2 * n + 2; ++i) cur[i] = fst[i], h[i] = -1;
	h[2 * n + 1] = 0;
	q[ed++] = 2 * n + 1;
	while (st < ed) {
		int u = q[st++];
		for (int i = fst[u]; i; i = nxt[i])
			if (f[i] > 0 && h[to[i]] == -1) {
				h[to[i]] = h[u] + 1;
				if (to[i] == 2 * n + 2) return true;
				q[ed++] = to[i];
			}
	}
	return false;
}
int dfs(int u, int fl) {
	if (u == 2 * n + 2 || !fl) return fl;
	int left = fl;
	for (int &i = cur[u], t; i; i = nxt[i])
		if (h[to[i]] == h[u] + 1 && (t = dfs(to[i], min(left, f[i]))) > 0) {
			f[i] -= t, f[i ^ 1] += t, left -= t;
			if (!left) break;
		}
	return fl - left;
}

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) {
		for (int j = 1; j <= n * 2 + 2; ++j)
			fst[j] = 0;
		static int s[105][105];
		tot = 1;
		for (int j = 1; j <= n; ++j) {
			for (int k = 1; k <= n; ++k)
				s[j][k] = 0;
			for (int k = i; k <= m; ++k)
				s[j][(A[j][k] - 1) / m + 1] = k;
			for (int k = 1; k <= n; ++k)
				if (s[j][k]) addedge(j, n + k, 1);
			addedge(n * 2 + 1, j, 1);
			addedge(n + j, n * 2 + 2, 1);
		}
		int res = 0;
		while (bfs()) res += dfs(2 * n + 1, 1 << 30);
		for (int j = 2; j <= tot; j += 2) {
			int u = to[j ^ 1], v = to[j];
			if (u >= 1 && u <= n && v > n && v <= 2 * n && !f[j])
				swap(A[u][i], A[u][s[u][v - n]]);
		}
	}
	static int B[105][105];
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			printf("%d%c", A[i][j], " \n"[j == m]), B[j][i] = A[i][j];
	for (int i = 1; i <= m; ++i) {
		sort(B[i] + 1, B[i] + 1 + n, [](int x, int y) { return (x - 1) / m < (y - 1) / m; });
		for (int j = 1; j <= n; ++j)
			A[j][i] = B[i][j];
	}
	for (int i = 1; i <= n; ++i)
		for (int j = 1; j <= m; ++j)
			printf("%d%c", A[i][j], " \n"[j == m]);
	return 0;
}