/* FILE: maxCommonSubsequence3Only2.cpp   last change: 23-Jan-2013   author: Romeo Rizzi
 * a solver for problem maxCommonSubsequence3 assuming the third input string is the concatenation of the first two (whence can be simply disregarded)
 */

#define NDEBUG   // NDEBUG definita nella versione che consegno
#include <cassert>
#ifndef NDEBUG
#  include <iostream>  // uso di cin e cout non consentito in versione finale
#endif
#include <fstream>

using namespace std;

const int MAX_N = 100;  // massima cardinalita' delle 3 stringhe in input
char s1[MAX_N], s2[MAX_N], s3[MAX_N]; // le 3 stringhe in input
int n1, n2, n3;  // lunghezza delle 3 stringhe in input

int opt[MAX_N +1][MAX_N +1];   /* opt[i][j] = the maximum length of a common subsequence of the two prefixes s1[0..i] and s2[0..j]. */

int main() {
  ifstream fin("input.txt"); assert( fin );
  fin >> n1 >> n2 >> n3;
  for(int i = 0; i < n1; i++)  fin >> s1[i];
  for(int i = 0; i < n2; i++)  fin >> s2[i];
  fin.close();
  for(int i = n1; i >= 0; i--)
    for(int j = n2; j >= 0; j--)
        if( (i == n1) || (j == n2) ) opt[i][j] = 0;
        else if( s1[i] != s2[j] ) opt[i][j] = max( opt[i+1][j], opt[i][j+1] );
        else // s1[i] == s2[j]
          opt[i][j] = 1 + opt[i+1][j+1];
  ofstream fout("output.txt");
  fout << opt[0][0] << endl;
  fout.close();
  return 0;
}