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>
#define pb push_back
#define pll pair <ll, ll>
#define mp make_pair
#define pyshnapyshnakaa ios_base :: sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define x first
#define y second
#pragma GCC optimize("O3")
// #pragma GCC optimize("Ofast")
// #pragma GCC optimize("unroll-loops")
#define plll pair <ll, pair <ll, ll>>
#define pllll pair <pair <ll, ll>, pair <ll, ll> >
#define psl pair <string, ll>
#define all(a) a.begin(), a.end()
#define vvl vector <vector <ll> >
 
typedef long long ll;
 
typedef long double ld;
 
using namespace std;
 
const ll maxn = 1e3 + 100;

ll n, m, k, t;

ll A[maxn][maxn];

ll geti(ll x) {
    return x / m;
}

void output() {
    ll q, w;
    for (q = 0; q < n; q++) {
        for (w = 0; w < m; w++) {
            cout << A[q][w] + 1 << " ";
        }
        cout << endl;
    }
}

vector <ll> V[maxn];

ll R[maxn];

set <ll> S[maxn];

bool used[maxn];

bool dfs(ll v) {
    used[v] = 1;
    for (auto v1 : V[v]) {
        if (R[v1] == -1) {
            R[v1] = v;
            return 1;
        }
    }
    for (auto v1 : V[v]) {
        if (!used[R[v1]]) {
            if (dfs(R[v1])) {
                R[v1] = v;
                return 1; 
            }
        }
    }
    return 0;
}

bool cmp(ll a, ll b) {
    return geti(a) < geti(b);
}

int main() {
    pyshnapyshnakaa;
    ll q, w, e, a, b, c;
    cin >> n >> m;
    vector <map <ll, vector <ll> > > M(n);
    for (q = 0; q < n; q++) {
        for (w = 0; w < m; w++) {
            cin >> A[q][w];
            A[q][w]--;
            M[q][geti(A[q][w])].pb(A[q][w]);
        }
    }
    for (q = 0; q < m; q++) {
        for (w = 0; w < n; w++) { // color
            V[w].clear();
            R[w] = -1;
            for (e = 0; e < n; e++) {
                if (!M[e][w].empty()) {
                    V[w].pb(e);
                }
            }
        }
        for (w = 0; w < n; w++) {
            for (e = 0; e < n; e++) {
                used[e] = 0;
            }
            dfs(w);
        }
        for (w = 0; w < n; w++) {
            ll c = R[w];
            A[w][q] = M[w][c].back();
            M[w][c].pop_back();
        }
    }
    output();
    for (q = 0; q < m; q++) {
        vector <ll> X;
        for (w = 0; w < n; w++) {
            X.pb(A[w][q]);
        }
        sort(all(X));
        for (w = 0; w < n; w++) {
            A[w][q] = X[w];
        }
    }
    output();
    return 0;
}