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:
/// awoooooo

# include <bits/stdc++.h>
# define sz(x) (int)((x).size())

using namespace std;

typedef long long ll;
typedef long double ld;
typedef pair < int, int > pii;

const int N = (int)2e5 + 5;
const int inf = (int)1e9 + 7;

const int D = 1000;

int n, m;
int a[D][D], b[D][D], u[D][D];
set < int > c[D];

int get(int x){
  return (x - 1) / m;
}

set < int > g[D];
int mt[D], used[D];

inline int dfs(int v){
  if(used[v]){
    return 0;
  }
  used[v] = 1;
  for(int i : g[v]){
    if(mt[i] == -1 || dfs(mt[i])){
      mt[i] = v;
      mt[v] = i;
      return 1;
    }
  }
  return 0;
}

void solve(int c){
  for(int i = 1; i <= n; ++i){
    g[i].clear();
  }
  for(int i = 1; i <= n; ++i){
    for(int j = c; j <= m; ++j){
      g[i].insert(get(a[i][j]) + n + 1);
    }
  }
  for(int i = 1; i <= 2 * n; ++i){
    mt[i] = -1;
    used[i] = 0;
  }
  for(int i = 1; i <= n; ++i){
    fill(used, used + D, 0);
    dfs(i);
  }
  for(int i = 1; i <= n; ++i){
    for(int j = c; j <= m; ++j){
      if(mt[i] - n - 1 == get(a[i][j])){
        swap(a[i][c], a[i][j]);
        break;
      }
    }
  }
}


int main(){
  cin >> n >> m;
  for(int i = 1; i <= n; ++i){
    for(int j = 1; j <= m; ++j){
      cin >> a[i][j];
    }
  }
  for(int i = 1; i <= m; ++i){
    solve(i);
  }
  for(int i = 1; i <= n; ++i){
    for(int j = 1; j <= m; ++j){
      printf("%d ", a[i][j]);
    }
    printf("\n");
  }
  for(int j = 1; j <= m; ++j){
    for(int i = 1; i <= n; ++i){
      b[get(a[i][j]) + 1][j] = a[i][j];
    }
  }
  for(int i = 1; i <= n; ++i){
    for(int j = 1; j <= m; ++j){
      printf("%d ", b[i][j]);
    }
    printf("\n");
  }



  return 0;
}