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<cstdio>
const int MAXN = 100;
const int MAXS = MAXN*MAXN;
struct edge{
	int to, num; edge *nxt;
}edges[MAXS + 5], *adj[MAXN + 5], *ecnt=&edges[0];
void addedge(int u, int v) {
	edge *p = (++ecnt);
	p->to = v, p->num = 0, p->nxt = adj[u], adj[u] = p;
//	printf("! %d %d\n", u, v);
}
int A[MAXN + 5][MAXN + 5];
int x[MAXS + 5], y[MAXS + 5], N, M, tot;
int vis[MAXN + 5], lnk[MAXN + 5], nw;
bool dfs(int x) {
	for(edge *p=adj[x];p;p=p->nxt)
		if( !p->num ) {
			if( !vis[p->to] ) {
				vis[p->to] = true;
				if( !lnk[p->to] || dfs(lnk[p->to]) ) {
					lnk[p->to] = x;
					return true;
				}
			}
		}
	return false;
}
int f[MAXS + 5];
int B[MAXN + 5][MAXN + 5], C[MAXN + 5][MAXN + 5];
int main() {
	scanf("%d%d", &N, &M);
	for(int i=1;i<=N;i++)
		for(int j=1;j<=M;j++) {
			scanf("%d", &A[i][j]);
			x[A[i][j]] = i, y[A[i][j]] = j;
		}
	tot = N*M;
	for(int i=1;i<=tot;i++)
		addedge(x[i], (i - 1) / M + 1);
	for(nw=1;nw<=M;nw++) {
		for(int i=1;i<=N;i++) lnk[i] = 0;
		for(int i=1;i<=N;i++) {
			for(int j=1;j<=N;j++)
				vis[j] = false;
			dfs(i);
		}
		for(int i=1;i<=N;i++) {
			for(edge *p=adj[i];p;p=p->nxt)
				if( !p->num ) {
					if( lnk[p->to] == i ) {
						p->num = nw;
						break;
					}
				}
		}
	}
	for(int i=1;i<=N*M;i++)
		B[x[i]][edges[i].num] = i;
	for(int i=1;i<=N;i++)
		for(int j=1;j<=M;j++)
			printf("%d%c", B[i][j], (j == M) ? '\n' : ' ');
	for(int i=1;i<=N*M;i++)
		C[(i - 1) / M + 1][edges[i].num] = i;
	for(int i=1;i<=N;i++)
		for(int j=1;j<=M;j++)
			printf("%d%c", C[i][j], (j == M) ? '\n' : ' ');
}