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;
bool vis[1005],used[505*505];
int p[505],N,M,a[505][505],g[505][505],qwq[505][505];vector<int> A[505],B[505];
int dfs(int now){
	for (int i=0;i<N;i++)
		if (g[now][i]&&!vis[i]){
			vis[i]=1;
			if (p[i]==-1||(dfs(p[i]))) {p[i]=now;return 1;}	
		}
		return 0;
}
int main(){
	scanf("%d%d",&N,&M);
	int x;
	for (int i=0;i<N;i++)
		for (int j=0;j<M;j++){
			scanf("%d",&a[i][j]);
			a[i][j]--;
		}
	for (int i=0;i<M;i++){//进行M-1次匹配
		for (int j=0;j<N;j++) 
			for (int k=0;k<N;k++) g[j][k]=0,qwq[j][k]=-1;
		for (int j=0;j<N;j++)
			for (int k=0;k<M;k++)
				if (!used[a[j][k]]){
					int x=a[j][k];
					g[j][x/M]=1;
					qwq[j][x/M]=x;
				}
		for (int j=0;j<N;j++) p[j]=-1;
		for (int j=0;j<N;j++){
			memset(vis,0,sizeof(vis));
			assert(dfs(j));//如果没有增广路了就退出 
		}
		for (int j=0;j<N;j++){
			int x=qwq[p[j]][j];
			assert(x!=-1);//找到与其匹配的点 
			used[x]=true;
			A[p[j]].push_back(x);
			B[i].push_back(x);
		}
	}
	for (int i=0;i<N;i++){ 
		for (int j=0;j<M;j++) printf("%d ",A[i][j]+1);
		printf("\n");
	}
	for (int i=0;i<M;i++) sort(B[i].begin(),B[i].end());
	for (int i=0;i<N;i++){
		for (int j=0;j<M;j++)
			printf("%d ",B[j][i]+1);
		printf("\n");
	}
	//puts("wqsb");
	return 0;
}