/*
1. 状态定义: 是A的子序列又是B的子序列的字符串长度最长是多少 -> f[i][j] 由A[1~i] 和 B[1~j] 字母组成的子序列的最大长度
2. 状态计算: 以最后一个位置划分, 为 4种情况:
    都不选  = f[i-1][j-1]
    选A[i]  = f[i][j-1] + 1
    选B[j]  = f[i-1][j] + 1
    都选    = f[i-1][j-1] + 1

3. 边界: f[~][~] = 0
*/

import java.util.*;

public class Main{

    void run(){
        int n = jin.nextInt();
        int m = jin.nextInt();
        String stringA = jin.next();
        String stringB = jin.next();
        int[][] f = new int[n+1][m+1];

        for (int i = 1; i <= n ; i ++){
            for (int j = 1; j <= m; j ++){
                f[i][j] = f[i-1][j-1];
                f[i][j] = Math.max(f[i][j], f[i-1][j]);
                f[i][j] = Math.max(f[i][j], f[i][j-1]);
                if (stringA.charAt(i-1) == stringB.charAt(j-1)) f[i][j] = Math.max(f[i][j], f[i-1][j-1] + 1);
            }
        }
        int res = f[n][m];
        System.out.println(res);
    }

    private Scanner jin = new Scanner(System.in);
    public static void main(String[] args) {new Main().run();}
}