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 long double ld;
typedef pair<int, int> ii;
typedef pair<int, ii> iii;
typedef pair<ii, int> ri3;
#define mp make_pair
#define pb push_back
#define fi first
#define sc second
#define SZ(x) (int)(x).size()
#define ALL(x) begin(x), end(x) 
#define REP(i, n) for (int i = 0; i < n; ++i) 
#define FOR(i, a, b) for (int i = a; i <= b; ++i)
#define RFOR(i, a, b) for (int i = a; i >= b; --i)

const int MX_N = 105;
const int MX_M = 105;

int N, M;
int B[MX_N][MX_M], C[MX_M][MX_N];

vector<int> al[2*MX_N];

struct graph {
    vector<int> al[2*MX_N];
    bool vis[2*MX_N];
    int match[2*MX_N];
    void reset() {
        FOR(i,1,2*N) al[i].clear();
    }
    int aug(int u) {
        if (vis[u]) return 0;
        vis[u] = 1;
        for (int v : al[u]) if (match[v] == -1 || aug(match[v])) {
            match[u] = v;
            match[v] = u;
            return 1;
        }
        return 0;
    }
    int mcbm() {
        memset(match,-1,sizeof match);
        int res = 0;
        FOR(i,1,N) if (match[i] == -1) {
            memset(vis,0,sizeof vis);
            res += aug(i);
        }
        return res;
    }
} g;

int main() {
    //freopen("in.txt", "r", stdin);
    ios::sync_with_stdio(false);
    cin.tie(0);

    cin >> N >> M;
    FOR(i,1,N){
        FOR(j,1,M){
            cin >> B[i][j];
        }
    }

    FOR(c,1,M){
        g.reset();
        FOR(i,1,N){
            FOR(j,c,M){
                int r = (B[i][j]+M-1)/M;
                g.al[r+N].push_back(i);
                g.al[i].push_back(r+N);
            }
        }
        assert(g.mcbm() == N);
        FOR(i,1,N){
            int r = g.match[i]-N;
            FOR(j,c,M) {
                if ((B[i][j]+M-1)/M == r) { swap(B[i][j],B[i][c]); break; }
            }
        }
        //FOR(i,1,N){
        //    cout << B[i][c] << ' ';
        //} cout << endl;
    }

    FOR(i,1,N){
        FOR(j,1,M){
            cout << B[i][j] << (j==M?'\n':' ');
            C[j][i] = B[i][j];
        }
    }

    FOR(j,1,M){
        sort(C[j]+1,C[j]+1+N);
    }

    FOR(i,1,N){
        FOR(j,1,M){
            cout << C[j][i] << (j==M?'\n':' ');
        }
    }
}