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;
using lint = long long;
using pi = pair<int, int>;
#define sz(v) ((int)(v).size())
#define all(v) (v).begin(), (v).end()
const int MAXN = 105;
const int MAXM = 10005;

struct edge_color{
	int n, m;
	int deg[2][MAXN];
	pi has[2][MAXN][MAXN];
	int color[MAXM];
	int c[2];

	void dfs(int x, int p) {
		auto i = has[p][x][c[!p]];
		if (has[!p][i.first][c[p]].second) dfs(i.first,!p);
		else has[!p][i.first][c[!p]] = pi(0,0);
		has[p][x][c[p]] = i;
		has[!p][i.first][c[p]] = pi(x,i.second);
		color[i.second] = c[p];
	}

	int solve(vector<pi> v, vector<int> &cv){
		m = sz(v);
		int ans = 0;
		for (int i=1;i<=m;i++) {
			int x[2];
			x[0] = v[i-1].first;
			x[1] = v[i-1].second;
			for (int d=0;d<2;d++) {
				deg[d][x[d]]+=1;
				ans = max(ans,deg[d][x[d]]);
				for (c[d]=1;has[d][x[d]][c[d]].second;c[d]++);
			}
			if (c[0]!=c[1]) dfs(x[1],1);
			for (int d=0;d<2;d++) has[d][x[d]][c[0]] = pi(x[!d],i);
			color[i] = c[0];
		}
		cv.resize(m);
		for(int i=1; i<=m; i++) cv[i-1] = color[i];
		return ans;
	}
}EC;

int n, m, a[MAXN][MAXN], b[MAXN][MAXN], idx[MAXN*MAXN];

int main(){
	auto print_grid = [&](int v[MAXN][MAXN]){
		for(int i=0; i<n; i++){
			for(int j=0; j<m; j++){
				printf("%d ", v[i][j]);
			}
			puts("");
		}
	};
	scanf("%d %d",&n,&m);
	for(int i=0; i<n; i++){
		for(int j=0; j<m; j++){
			scanf("%d",&a[i][j]);
		}
	}
	vector<pi> v;
	for(int i=0; i<n; i++){
		for(int j=0; j<m; j++){
			int col = (a[i][j] - 1) / m;
			idx[a[i][j]] = sz(v);
			v.emplace_back(i+1, col+1);
		}
	}
	vector<int> color;
	assert(EC.solve(v, color) == m);
	for(int i=1; i<=n*m; i++){
		int iv = idx[i];
		b[v[iv].first-1][color[iv] - 1] = i;
	}
	print_grid(b);
	for(int i=0; i<m; i++){
		vector<int> v;
		for(int j=0; j<n; j++) v.push_back(b[j][i]);
		sort(all(v));
		for(int j=0; j<n; j++) b[j][i] = v[j];
	}
	print_grid(b);
}