/*
 * @Author: YMYS
 * @Date: 2024-02-25 21:38:22
 * @LastEditTime: 2024-12-22 11:10:03
 * @FilePath: \VScode-C&C++-Coding\Acwing\算法基础课\5.第五章\2.线性DP\4.最长公共子序列.cpp
 * @URL:https://www.acwing.com/problem/content/899/
 * @Description: 897. 最长公共子序列
 * 
 * 题解推荐：https://www.acwing.com/solution/content/8111/
 * 
 * 推荐的blibli视频：https://www.bilibili.com/video/BV1ey4y1d7oD/
 * 
 * 最长公共子序列：可以断开
 * 最长公共子串：不能断开
 * eg：x：ABCBDAB；y：BDCABC
 * 
 */
#include<bits/stdc++.h>

using namespace std;

const int N = 1010;

int n,m;
char a[N],b[N];
int f[N][N];

signed main()
{
    cin>>n>>m>>a+1>>b+1;

    for(int i=1;i<=n;i++){
        for(int j=1;j<=m;j++){
            f[i][j] = max(f[i-1][j], f[i][j-1]);
            if(a[i] == b[j]) f[i][j] = f[i-1][j-1] + 1;
        }
    }

    cout << f[n][m] << endl;

    return 0;
}