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>
#define EB emplace_back

typedef std::pair <int, int> pr;
typedef std::vector <pr> vecpr;
const int N = 108, N2 = 10054;

int R, C;
int A1[N2], A2[N2], ans1[N][N], ans2[N][N];
int matL[N], matR[N];
bool used[N];
vecpr G[N];

bool dfs(int x) {
	int y;
	for (const pr &p : G[x])
		if (!used[y = p.first])
			if (used[y] = true, !~matL[y] || dfs(matL[y]))
				return matL[y] = x, matR[x] = y, true;
	return false;
}

int main() {
	int i, j, x, n, match;
	scanf("%d%d", &R, &C);
	for (i = 0; i < R; ++i) {
		for (j = 0; j < C; ++j) scanf("%d", &x), A1[--x] = i;
		std::fill(A2 + i * C, A2 + (i + 1) * C, i);
	}
	for (i = 0; i < R * C; ++i) G[ A1[i] ].EB(A2[i], i + 1);
	for (match = 0; match < C; ++match) {
		memset(matL, -1, R << 2), memset(matR, -1, R << 2);
		for (i = 0; i < R; ++i) memset(used, false, R), assert( dfs(i) );
		for (i = 0; i < R; ++i)
			for (n = G[i].size(), j = 0; j < n; ++j)
				if (G[i][j].first == matR[i]) {
					ans1[i][match] = ans2[ matR[i] ][match] = G[i][j].second;
					G[i].erase(G[i].begin() + j); break;
				}
	}
	for (i = 0; i < R; ++i) for (j = 0; j < C; ++j) printf("%d%c", ans1[i][j], j == C - 1 ? 10 : 32);
	for (i = 0; i < R; ++i) for (j = 0; j < C; ++j) printf("%d%c", ans2[i][j], j == C - 1 ? 10 : 32);
	return 0;
}