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>
#pragma GCC optimize ("O2,unroll-loops")
//#pragma GCC optimize("no-stack-protector,fast-math")
//#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")

using namespace std;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<pii, int> piii;
typedef pair<ll, ll> pll;
#define debug(x) cerr<<#x<<'='<<(x)<<endl;
#define debugp(x) cerr<<#x<<"= {"<<(x.first)<<", "<<(x.second)<<"}"<<endl;
#define debug2(x, y) cerr<<"{"<<#x<<", "<<#y<<"} = {"<<(x)<<", "<<(y)<<"}"<<endl;
#define debugv(v) {cerr<<#v<<" : ";for (auto x:v) cerr<<x<<' ';cerr<<endl;}
#define all(x) x.begin(), x.end()
#define pb push_back
#define kill(x) return cout<<x<<'\n', 0;

const ld eps=1e-7;
const int inf=1000000010;
const ll INF=10000000000000010LL;
const int mod=1000000007;
const int N=105;

int n, m, k, u, v, x, y, t, a, b, ans;
int A[N][N];
int B[N][N];
int C[N][N];
int match[N], tmp[N];
bool mark[N], ok[N];
bool G[N][N];

bool dfs(int node){
	mark[node]=1;
	for (int v=1; v<=n; v++) if (G[node][v]){
		if (!match[v] || (!mark[match[v]] && dfs(match[v]))){
			match[v]=node;
			return 1;
		}
	}
	return 0;
}

int main(){
	ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
	//freopen("input.txt", "r", stdin);
	//freopen("output.txt", "w", stdout);
	cin>>n>>m;
	for (int i=1; i<=n; i++) for (int j=1; j<=m; j++) cin>>A[i][j];
	for (int j=1; j<=m; j++){
		memset(G, 0, sizeof(G));
		for (int x=1; x<=n; x++) for (int y=1; y<=m; y++) if (A[x][y]) G[x][(A[x][y]+m-1)/m]=1;
		memset(match, 0, sizeof(match));
		memset(ok, 0, sizeof(ok));
//		debug("start")
		int flag=1;
		while (flag--){
			memset(mark, 0, sizeof(mark));
			for (int i=1; i<=n; i++) if (!mark[i] && !ok[i] && dfs(i)) flag=ok[i]=1;
		}
		for (int i=1; i<=n; i++){
			assert(match[i]);
//			debug2(i, match[i])
			int x=match[i];
			for (int y=1; y<=m; y++) if ((A[x][y]+m-1)/m==i){
				B[x][j]=A[x][y];
				A[x][y]=0;
				break ;
			}
		}
	}
	for (int j=1; j<=m; j++){
		for (int i=1; i<=n; i++) tmp[i]=B[i][j];
		sort(tmp+1, tmp+n+1);
		for (int i=1; i<=n; i++) C[i][j]=tmp[i];
	}
	for (int i=1; i<=n; i++) for (int j=1; j<=m; j++) cout<<B[i][j]<<" \n"[j==m];
	for (int i=1; i<=n; i++) for (int j=1; j<=m; j++) cout<<C[i][j]<<" \n"[j==m];
	
	return 0;
}