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:
#pragma GCC optimize("O3")
#include <bits/stdc++.h>
using namespace std;
typedef long double ld;
typedef long long ll;
const int maxN = 105;
int n, m;
int id[maxN * maxN];
vector < int > a[maxN];
vector < int > g[maxN];
bool was[maxN][maxN];
int mt[maxN];
bool used[maxN];
int b[maxN][maxN];
int c[maxN][maxN];
bool kuhn(int v) {
    if (used[v]) return false;
    used[v] = true;
    for (int to : g[v]) {
        if (mt[to] == -1 || kuhn(mt[to])) {
            mt[to] = v;
            return true;
        }
    }
    return false;
}
int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    //freopen("input.txt", "r", stdin);
    cin >> n >> m;
    for (int i = 1; i <= n; i++) {
        a[i].resize(m);
        for (int& v : a[i]) {
            cin >> v;
            id[v] = (v - 1) / m + 1;
        }
    }
    for (int it = 0; it < m; it++) {
        memset(was, 0, sizeof was);
        memset(used, 0, sizeof used);
        memset(mt, -1, sizeof mt);
        for (int i = 1; i <= n; i++) {
            g[i].clear();
            for (int v : a[i]) {
                if (!was[i][id[v]]) {
                    g[i].push_back(id[v]);
                    was[i][id[v]] = true;
                }
            }
        }
        for (int i = 1; i <= n; i++) {
            memset(used, 0, sizeof used);
            assert(kuhn(i));
        }
        for (int i = 1; i <= n; i++) {
            assert(mt[i] != -1);
            for (int j = 0; j < a[mt[i]].size(); j++) {
                if (id[a[mt[i]][j]] == i) {
                    b[mt[i]][it + 1] = a[mt[i]][j];
                    a[mt[i]].erase(a[mt[i]].begin() + j);
                    break;
                }
            }
        }
    }
    for (int col = 1; col <= m; col++) {
        for (int row = 1; row <= n; row++) {
            c[id[b[row][col]]][col] = b[row][col];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cout << b[i][j] << " ";
        }
        cout << '\n';
    }

    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            cout << c[i][j] << " ";
        }
        cout << '\n';
    }
    return 0;
}