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;
typedef long long LL;
typedef pair<int, int> pii;
#ifdef DEBUG
#define display(x) cerr << #x << " = " << x << endl;
#define displaya(a, st, n)\
	{cerr << #a << " = {";\
	for(int qwq = (st); qwq <= (n); ++qwq) {\
		if(qwq == (st)) cerr << a[qwq];\
		else cerr << ", " << a[qwq];\
	} cerr << "}" << endl;}
#define displayv(v) displaya(v, 0, (int)(v).size() - 1)
#define eprintf(...) fprintf(stderr, __VA_ARGS__)
#else
#define display(x) ;
#define displaya(a, st, n) ;
#define displayv(v) ;
#define eprintf(...) if(0) fprintf(stderr, "...")
#endif
template<typename T> bool chmin(T &a, const T &b) { return a > b ? a = b, true : false; }
template<typename T> bool chmax(T &a, const T &b) { return a < b ? a = b, true : false; }

const int maxN = 100 + 5;
int n, m;
int G[maxN][maxN];
vector<int> q[maxN][maxN];
int link[maxN];
bool vis[maxN];

int x[maxN * maxN], y[maxN * maxN], col[maxN * maxN];

int dfs(int u) {
	for(int v = 1; v <= n; ++v) if(G[u][v] && !vis[v]) {
		vis[v] = true;
		if(link[v] == 0) link[v] = u;
		else if(dfs(link[v])) {
			link[v] = u;
		} else {
			continue;
		}
		return true;
	}
	return false;
}

int main() {
#ifndef LOCAL
	ios::sync_with_stdio(false);
	cin.tie(0); cout.tie(0);
#endif
	cin >> n >> m;
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j) {
			int p; cin >> p;
			x[p] = i;
			y[p] = (p - 1) / m + 1;
			G[x[p]][y[p]]++;
			q[x[p]][y[p]].push_back(p);
		}
	for(int i = 1; i <= m; ++i) {
		memset(link, 0, sizeof(link));
		for(int u = 1; u <= n; ++u) {
			memset(vis, 0, sizeof(vis));
			if(!dfs(u)) assert(false);
		}
		for(int v = 1; v <= n; ++v) {
			int u = link[v];
			col[q[u][v].back()] = i;
			q[u][v].pop_back();
			G[u][v]--;
		}
	}
	for(int i = 1; i <= n * m; ++i) {
		G[x[i]][col[i]] = i;
	}
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			printf("%d%c", G[i][j], " \n"[j == m]);
	for(int i = 1; i <= n * m; ++i) {
		G[y[i]][col[i]] = i;
	}
	for(int i = 1; i <= n; ++i)
		for(int j = 1; j <= m; ++j)
			printf("%d%c", G[i][j], " \n"[j == m]);
	return 0;
}