#include <bits/stdc++.h>
#define pb push_back
using namespace std;
ifstream fi("cuplaj.in");
ofstream fo("cuplaj.out");

const int NMAX = 5005;

int F[NMAX][NMAX], C[NMAX][NMAX];
vector <int> G[NMAX];
int n, m, e;
queue <int> Q;
bool viz[NMAX];
int p[NMAX];
vector < pair<int, int> > muchiiRez;

int bfs()
{
    memset(viz, 0, sizeof(viz));
    memset(p, 0, sizeof(p));

    while (!Q.empty())
        Q.pop();

    Q.push(1);
    viz[1] = 1;

    while (!Q.empty())
    {
        int curr = Q.front();
        Q.pop();

        for (auto v: G[curr])
        {
            if (F[curr][v] != C[curr][v] && !viz[v])
            {
                p[v] = curr;
                Q.push(v);
                viz[v] = 1;
                if (v == n + m + 2)
                    return 1;
            }
        }
    }
    return viz[n + m + 2];
}

void dfs(int nod)
{
    viz[nod] = 1;
    for (auto v: G[nod])
        if (!viz[v] && F[nod][v] == C[nod][v])
        {
            if (nod != 1 && v != n + m + 2)
                muchiiRez.pb({nod, v});
            dfs(v);
        }
}

int main()
{
    fi >> n >> m >> e;

    for (int i = 1; i <= e; i++)
    {
        int u, v;
        fi >> u >> v;
        v += n;
        u++;
        v++;
        G[u].pb(v);
        G[v].pb(u);
        C[u][v] = 1;
    }

    for (int i = 2; i <= n + 1; i++)
    {
        G[1].pb(i);
        G[i].pb(1);
        C[1][i] = 1;
    }

    for (int i = n + 2; i <= n + m + 1; i++)
    {
        G[i].pb(n + m + 2);
        G[n + m + 2].pb(i);
        C[i][n + m + 2] = 1;
    }

    int rez = 0;

    while (bfs())
    {
        int minim = 100;

        for (int i = n + m + 2; i != 1; i = p[i])
        {
            minim = min(minim, C[p[i]][i] - F[p[i]][i]);
        }

        for (int i = n + m + 2; i != 1; i = p[i])
        {
            F[p[i]][i] += minim;
            F[i][p[i]] -= minim;
        }

        rez += minim;
    }

    fo << rez << "\n";

    for (int i = 2; i <= n + 1; i++)
    {
        for (auto v: G[i])
        {
            if (C[i][v] == F[i][v] && C[i][v] == 1)
                fo << i - 1 << " " << v - (n + 1) << "\n";
        }
    }

    return 0;
}