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;

typedef long long ll;
typedef unsigned int ui;
typedef pair<int, int> pi;

#define fi first
#define se second

int read() {
  int res = 0, w = 1; char c = getchar();
  while (!isdigit(c) && c != '-') c = getchar();
  if (c == '-') c = getchar(), w = -1;
  while (isdigit(c)) res = (res << 1) + (res << 3) + c - '0', c = getchar();
  return res * w;
}

const int N = 110;

int n, m, a[N][N], b[N][N], c[N][N];

struct Edge {
  int u, v, nxt, id, mark;
} E[N * N];

int head[N], tot = 0, vis[N], match[N];

void addedge(int u, int v, int id) {
  E[++tot] = (Edge) {u, v, head[u], id, 0};
  head[u] = tot;
}

bool dfs(int u) {
  for (int i = head[u]; i; i = E[i].nxt) if (!E[i].mark) {
    int v = E[i].v;
    if (vis[v]) continue;
    vis[v] = 1;
    if (!match[v] || dfs(E[match[v]].u)) {
      match[v] = i;
      return 1;
    }
  }
  return 0;
}

int main() {
  n = read(), m = read();
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      a[i][j] = read();
      addedge(i, (a[i][j] - 1) / m + 1, a[i][j]);
    }
  }
  for (int i = 1; i <= m; i++) {
    memset(match, 0, sizeof(match));
    for (int j = 1; j <= n; j++) {
      memset(vis, 0, sizeof(vis));
      dfs(j);
    } 
    for (int j = 1; j <= n; j++) {
      b[E[match[j]].u][i] = E[match[j]].id;
      c[E[match[j]].v][i] = E[match[j]].id;
      E[match[j]].mark = 1;
    }
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      printf("%d ", b[i][j]);
    }
    puts("");
  }
  for (int i = 1; i <= n; i++) {
    for (int j = 1; j <= m; j++) {
      printf("%d ", c[i][j]);
    }
    puts("");
  }
  return 0;
}