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;
#define rep(i,a,b) for (int i = (a); i <= (b); ++ i)
#define rrp(i,a,b) for (int i = (a); i >= (b); -- i)
#define gc() getchar()
#define fir first
#define sec second
typedef pair<int,int> pii;
typedef long double db;
typedef long long ll;
template <typename tp>
inline void read(tp& x) {
  x = 0; char tmp; bool key = 0;
  for (tmp = gc(); !isdigit(tmp); tmp = gc())
    key = (tmp == '-');
  for (; isdigit(tmp); tmp = gc())
    x = (x << 3) + (x << 1) + (tmp ^ '0');
  if (key) x = -x;
}
template <typename tp>
inline void ckmn(tp& x,tp y) {
  x = x < y ? x : y;
}
template <typename tp>
inline void ckmx(tp& x,tp y) {
  x = x < y ? y : x;
}

const int N = 110;
int n, m, match[N], vis[N];
pii mat[N][N], res[N][N], res2[N][N];
vector<int> vec[N][N];
bool dfs(int pos) {
  rep (i, 1, n) if (vec[pos][i].size() > 0) {
    if (vis[i] == 1) continue;
    vis[i] = 1;
    if (!match[i] || dfs(match[i])) {
      match[i] = pos;
      return true;
    }
  }
  return false;
}
void doit() {
  rep (i, 1, n) match[i] = 0;
  rep (i, 1, n) {
    memset(vis, 0, sizeof vis);
    dfs(i);
  }
}
int main() {
  read(n), read(m);
  rep (i, 1, n) {
    int x;
    rep (j, 1, m) {
      read(x);
      mat[i][j] = pii((x - 1) / m + 1, (x - 1) % m + 1);
      vec[i][mat[i][j].fir].push_back(mat[i][j].sec);
    }
  }
  rep (j, 1, m) {
    doit();
    rep (i, 1, n) {
      int p = match[i];
      res[p][j] = pii(i, vec[p][i].back());
      vec[p][i].pop_back();
      res2[i][j] = res[p][j];
    }
  }
  rep (i, 1, n) {
    rep (j, 1, m) {
      printf("%d ", (res[i][j].fir-1) * m + res[i][j].sec);
    }
    puts("");
  }
  rep (i, 1, n) {
    rep (j, 1, m) {
      printf("%d ", (res2[i][j].fir-1) * m + res2[i][j].sec);
    }
    puts("");
  }
  return 0;
}