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 rep(i, a, b) for (int i = a; i <= b; i++)
#define per(i, a, b) for (int i = a; i >= b; i--)
using namespace std;

typedef unsigned long long ull;
typedef pair <int, int> pii;
typedef long long ll;

template <typename _T>
inline void read(_T &f) {
	f = 0; _T fu = 1; char c = getchar();
	while (c < '0' || c > '9') { if (c == '-') { fu = -1; } c = getchar(); }
	while (c >= '0' && c <= '9') { f = (f << 3) + (f << 1) + (c & 15); c = getchar(); }
	f *= fu;
}

template <typename T>
void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x < 10) putchar(x + 48);
	else print(x / 10), putchar(x % 10 + 48);
}

template <typename T>
void print(T x, char t) {
	print(x); putchar(t);
}

const int N = 105;

vector <int> a[N][N];
int match[N], vis[N], adj[N][N], b[N][N], c[N][N];
int n, m;

int dfs1(int u) {
	for (int i = 1; i <= n; i++) {
		if (adj[u][i] && !vis[i]) {
			vis[i] = 1;
			if (!match[i] || dfs1(match[i])) {
				match[i] = u;
				return 1;
			}
		}
	}
	return 0;
}

int main() {
	read(n); read(m);
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			int x; read(x);
			a[i][(x - 1) / m + 1].push_back(x);
		}
	}
	for (int j = 1; j <= m; j++) {
		memset(match, 0, sizeof(match));
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= n; j++) {
				if (a[i][j].size()) adj[i][j] = 1;
				else adj[i][j] = 0;
			}
		}
		for (int i = 1; i <= n; i++) {
			memset(vis, 0, sizeof(vis));
			dfs1(i);
		}
		for (int i = 1; i <= n; i++) {
			int id = match[i];
			b[id][j] = a[id][i].back();
			a[id][i].pop_back();
		}
	}
	for (int j = 1; j <= m; j++) {
		for (int i = 1; i <= n; i++) c[i][j] = b[i][j];
		for (int i = 1; i <= n; i++) {
			for (int k = n; k > i; k--) {
				if (c[k][j] < c[k - 1][j]) {
					swap(c[k][j], c[k - 1][j]);
				}
			}
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			print(b[i][j], j == m ? '\n' : ' ');
		}
	}
	for (int i = 1; i <= n; i++) {
		for (int j = 1; j <= m; j++) {
			print(c[i][j], j == m ? '\n' : ' ');
		}
	}
	return 0;
}