//cu matrice de adiacenta
#include<fstream>
#include<cstring>
using namespace std;
ifstream fin("cuplaj.in");
ofstream fout("cuplaj.out");

struct muchie {int x,y;};
int C[10001][10001];//matricea capacitatilor
int t1,t2,n;//cardinalele si nr total de noduri
int T[10001],X[10001];//vector TATA din BFS, coada din BFS
muchie S[10001];//solutia
int sk;//lungimea solutiei
 
int BFS() //returneaza 1 daca gaseste drum de crestere de la 0 la n
{
    int s=1,d=1,j;
    memset(T,0,sizeof(T));
    memset(X,0,sizeof(X));
    X[1]=0;T[0]=-1;
    while(s<=d)
    {
        for(j=1;j<=n;j++)
            if(T[j]==0 && C[X[s]][j]>0)
            {
                X[++d]=j;
                T[j]=X[s];
                if(j==n)
                    return 1;
            }
        s++;
    }
    return 0;
}
 
void cuplaj_maxim()
{
    int j;
    while(BFS())//cat timp mai gaseste drumuri de crestere
      {
                j=n;
                while(j!=0)//inverseaza arcele
                    {   C[T[j]][j]=0;
                        C[j][T[j]]=1;
                        j=T[j];
                    }
      }
}
 
 
int main()
{
    int i,x,y,m;
    fin>>t1>>t2>>m;
    n=t1+t2;
    for(i=1;i<=m;++i)
    {
        fin>>x>>y;
        C[x][t1+y]=1;
    }
    for(i=1;i<=t1;i++) C[0][i]=1;
    n++;
    for(i=1;i<=t2;i++) C[t1+i][n]=1;
    cuplaj_maxim();
    for(i=1;i<=n;i++)
        if(C[n][i]==1) S[++sk].y=i;
    for(i=1;i<=sk;i++)
            for(int j=1;j<=n;j++)
                 if(C[S[i].y][j]) S[i].x=j;
    fout<<sk<<endl;
    for(i=1;i<=sk;i++) fout<<S[i].x<<" "<<S[i].y-t1<<endl;
    return 0;
}

// cu listele de adiacenta
#include<fstream>
#include<cstring>
#include <vector>
using namespace std;
ifstream fin("cuplaj.in");
ofstream fout("cuplaj.out");
struct muchie {int x,y;};
int t1,t2,n;//cardinalele si nr total de noduri
int T[10001],X[10001];//vector TATA din BFS, coada din BFS
muchie S[10001];//solutia
int sk;//lungimea solutiei
vector<int> V[10001];//listele de adiacenta
int BFS() //returneaza 1 daca gaseste drum de crestere de la 0 la n
{
    int s=1,d=1,j;
    memset(T,0,sizeof(T));
    memset(X,0,sizeof(X));
    X[1]=0;T[0]=-1;
    while(s<=d)
    {
        for(int i=0;i<V[X[s]].size();i++)
        {
            j=V[X[s]][i];
            if(T[j]==0)
            {
                X[++d]=j;
                T[j]=X[s];
                if(j==n)
                    return 1;
            }
        }
        s++;
    }
    return 0;
}
 
void cuplaj_maxim()
{
    int j;
    while(BFS())//cat timp mai gaseste drumuri de crestere
      {
                j=n;
                while(j!=0)//inverseaza arcele
                    {   for(int i=0;i<V[T[j]].size();i++)
                           if(V[T[j]][i]==j) V[T[j]].erase(V[T[j]].begin()+i);
                        V[j].push_back(T[j]);
                        j=T[j];
                    }
      }
}
 
 
int main()
{
    int i,x,y,m;
    fin>>t1>>t2>>m;
    n=t1+t2;
    for(i=1;i<=t1;i++) V[0].push_back(i);
    for(i=1;i<=m;++i)
    {
        fin>>x>>y;
        V[x].push_back(y+t1);
    }
    n++;
    for(i=1;i<=t2;i++) V[t1+i].push_back(n);
 
    cuplaj_maxim();
 
    for(i=0;i<V[n].size();i++)
        S[++sk].y=V[n][i];
    for(i=1;i<=sk;i++)
           for(int j=0;j<V[S[i].y].size();j++)
                S[i].x=V[S[i].y][j];
    fout<<sk<<endl;
    for(i=1;i<=sk;i++) fout<<S[i].x<<" "<<S[i].y-t1<<endl;
    return 0;
}