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 FOR(i, a, b) for(int i = a; i < b; ++i)
#define REP(i, n) FOR(i, 0, n)
#define _ << " " <<
#define pb push_back
#define fi first
#define se second

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

const int mod = 998244353;
int add(int x, int y) {x += y; if(x >= mod) return x - mod; return x;}
int sub(int x, int y) {x -= y; if(x < 0) return x + mod; return x;}
int mul(int x, int y) {return (ll) x * y % mod;}

const int MAXN = 105;

bool seen[MAXN];
int matchR[MAXN];

int n, m;
bool e[MAXN][MAXN];

bool f(int u) {
  REP(v, n) {
    if(!e[u][v] || seen[v]) continue;
    seen[v] = true;

    if(matchR[v] == -1 || f(matchR[v])) {
      matchR[v] = u;
      return true;
    }
  }
  return false;
}

int a[MAXN][MAXN];
int sol1[MAXN][MAXN];

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);

  cin >> n >> m;
  REP(i, n) REP(j, m) {
    cin >> a[i][j];
    a[i][j] --;
  }

  REP(col, m) {
    memset(e, 0, sizeof e);
    REP(i, n) matchR[i] = -1;

    REP(i, n) REP(j, m) {
      if(a[i][j] != -1) {
        e[a[i][j] / m][i] = 1;
      }
    }

    REP(i, n) {
      memset(seen, 0, sizeof seen);
      f(i);
    }

    REP(i, n) REP(j, m) {
      if(a[i][j] != -1 && a[i][j] / m == matchR[i]) {
        sol1[i][col] = a[i][j];
        a[i][j] = -1;
        break;
      }
    }
  }

  REP(i, n) {
    REP(j, m) cout << sol1[i][j] + 1 << " ";
    cout << endl;
  }

  REP(i, n) {
    REP(j, m) {
      REP(k, n) {
        if(sol1[k][j] / m == i) cout << sol1[k][j] + 1 << " ";
      }
    }
    cout << endl;
  }
}