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;

const int maxn = 110;
int n, m, a[maxn][maxn], b[maxn][maxn], c[maxn][maxn], lnk[maxn], vis[maxn];
vector<int> G[maxn];
int id(int i, int j) { return (i - 1) * m + j; }

bool match(int v) {
    for (int i = 0, u; i < G[v].size(); i++) {
        if (vis[u = G[v][i]]) continue; vis[u] = 1;
        if (!lnk[u] || match(lnk[u])) { lnk[u] = v; return 1; }
    }
    return 0;
}

int main() {
    scanf("%d %d", &n, &m);
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            scanf("%d", &a[i][j]);
        }
    }
    for (int i = 1; i <= m; i++) {
        for (int j = 1; j <= n; j++) {
            G[j].clear();
            for (int k = 1; k <= m; k++) {
                if (!b[j][k]) G[j].push_back((a[j][k] - 1) / m + 1);
            }
        }
        memset(lnk, 0, sizeof(lnk));
        for (int j = 1; j <= n; j++) {
            memset(vis, 0, sizeof(vis)), match(j);
        }
        for (int j = 1; j <= n; j++) {
            for (int k = 1; k <= m; k++) {
                int x = (a[j][k] - 1) / m + 1;
                if (!b[j][k] && lnk[x] == j) { b[j][k] = i; break; }
            }
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            c[b[i][j]][i] = a[i][j];
        }
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            printf("%d%c", c[j][i], " \n"[j == m]);
        }
    }
    for (int i = 1; i <= m; i++) {
        sort(c[i] + 1, c[i] + n + 1);
    }
    for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= m; j++) {
            printf("%d%c", c[j][i], " \n"[j == m]);
        }
    }
    return 0;
}