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>

struct vertex {
	int estart,estart2;
	int dist;
} v[203];
struct edge {
	int enext,to;
	int cap;
} e[40003];int en;
void addedge(int a,int b,int cap)
{
	e[en].enext = v[a].estart;e[en].to = b;e[en].cap = cap;v[a].estart = en++;
	e[en].enext = v[b].estart;e[en].to = a;e[en].cap = 0;v[b].estart = en++;
}
int dfs(int i,int cap)
{
	if(i == 1) return cap;
	int flow = 0;
	for(int &j = v[i].estart2;cap > 0 && j != -1;j = e[j].enext) {
		int to = e[j].to;
		if(v[to].dist != v[i].dist + 1) continue;
		int val = dfs(to,std::min(cap,e[j].cap));
		flow += val;cap -= val;e[j].cap -= val;e[j^1].cap += val;
	}
	return flow;
}
int n,m;
int a1[103][103],a2[103][103];
std::pair<int,int> pos[20003];
int main()
{
	scanf("%d%d",&n,&m);
	int vn = 2+n*2;
	for(int i = 0;i < vn;++i) {
		v[i].estart = -1;
	}
	for(int i = 0;i < n;++i) {
		for(int j = 0;j < m;++j) {
			int val;
			scanf("%d",&val);--val;
			pos[val] = std::make_pair(i,j);
		}
	}
	for(int i = 0;i < n*m;++i) {
		addedge(2+pos[i].first,2+n+i/m,1);
	}
	assert(en == 2*n*m);
	for(int i = 0;i < n;++i) {
		addedge(0,2+i,0);
		addedge(2+n+i,1,0);
	}
	for(int k = 0;k < m;++k) {
		// get exactly one matching
		for(int i = 0;i < n;++i) {
			++e[2*n*m+4*i].cap;
			++e[2*n*m+4*i+2].cap;
		}
		int flow = 0;
		while(flow < n) {
			// dinic
			for(int i = 0;i < vn;++i) {
				v[i].estart2 = v[i].estart;
				v[i].dist = -1;
			}
			v[0].dist = 0;
			std::queue<int> q;q.push(0);
			while(!q.empty()) {
				int cur = q.front();q.pop();
				for(int j = v[cur].estart;j != -1;j = e[j].enext) {
					if(e[j].cap == 0) continue;
					int to = e[j].to;
					if(v[to].dist == -1) {
						v[to].dist = v[cur].dist + 1;
						q.push(to);
					}
				}
			}
			if(v[1].dist == -1) break;
			flow += dfs(0,n);
		}
		assert(flow == n);
		// get result
		for(int i = 0;i < n*m;++i) {
			if(e[2*i+1].cap > 0) {
				e[2*i+1].cap = 0;
				a1[pos[i].first][k] = i;
				a2[i/m][k] = i;
			}
		}
	}
	for(int i = 0;i < n;++i) {
		for(int j = 0;j < m;++j) printf("%d ",a1[i][j]+1);
		printf("\n");
	}
	for(int i = 0;i < n;++i) {
		for(int j = 0;j < m;++j) printf("%d ",a2[i][j]+1);
		printf("\n");
	}
}