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 <cstdio>

const int MN = 105;

int N, M, A[MN][MN], B[MN][MN];

int cho[MN], mch[MN], vis[MN], v;
bool DFS(int i) {
	for (int j = 1; j <= M; ++j) if (A[i][j]) {
		int x = (A[i][j] - 1) / M + 1;
		if (vis[x] != v) {
			vis[x] = v;
			if (!mch[x] || DFS(mch[x])) return cho[i] = j, mch[x] = i, 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 j = 1; j <= M; ++j) {
		for (int i = 1; i <= N; ++i) v = (i - 1) * M + j, DFS(i);
		for (int i = 1; i <= N; ++i) B[i][j] = A[i][cho[i]], A[i][cho[i]] = 0, mch[i] = 0;
	}
	for (int i = 1; i <= N; ++i, puts(""))
		for (int j = 1; j <= M; ++j)
			printf("%d ", B[i][j]);
	for (int j = 1; j <= M; ++j) {
		static int per[MN];
		for (int i = 1; i <= N; ++i) per[(B[i][j] - 1) / M + 1] = B[i][j];
		for (int i = 1; i <= N; ++i) B[i][j] = per[i];
	}
	for (int i = 1; i <= N; ++i, puts(""))
		for (int j = 1; j <= M; ++j)
			printf("%d ", B[i][j]);
	return 0;
}