D. Catching Cheaters

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该博客讨论了一种计算两个字符串子序列相似度的方法。给定两个字符串，定义它们的相似度为4倍的最长公共子序列长度减去两字符串长度之和。问题要求找到所有字符串子序列对的最大相似度。提供的代码实现了一个动态规划解决方案来求解这个问题，并给出了几个测试用例以说明其工作原理。

https://codeforces.ml/contest/1447/problem/D

You are given two strings AA and BB representing essays of two students who are suspected cheaters. For any two strings CC, DD we define their similarity score S(C,D)S(C,D) as 4⋅LCS(C,D)−|C|−|D|4⋅LCS(C,D)−|C|−|D|, where LCS(C,D)LCS(C,D) denotes the length of the Longest Common Subsequence of strings CC and DD.

You believe that only some part of the essays could have been copied, therefore you're interested in their substrings.

Calculate the maximal similarity score over all pairs of substrings. More formally, output maximal S(C,D)S(C,D) over all pairs (C,D)(C,D), where CC is some substring of AA, and DD is some substring of BB.

If XX is a string, |X||X| denotes its length.

A string aa is a substring of a string bb if aa can be obtained from bb by deletion of several (possibly, zero or all) characters from the beginning and several (possibly, zero or all) characters from the end.

A string aa is a subsequence of a string bb if aa can be obtained from bb by deletion of several (possibly, zero or all) characters.

Pay attention to the difference between the substring and subsequence, as they both appear in the problem statement.

You may wish to read the Wikipedia page about the Longest Common Subsequence problem.

Input

The first line contains two positive integers nn and mm (1≤n,m≤50001≤n,m≤5000) — lengths of the two strings AA and BB.

The second line contains a string consisting of nn lowercase Latin letters — string AA.

The third line contains a string consisting of mm lowercase Latin letters — string BB.

Output

Output maximal S(C,D)S(C,D) over all pairs (C,D)(C,D), where CC is some substring of AA, and DD is some substring of BB.

Examples

input

Copy

4 5
abba
babab

output

Copy

input

Copy

8 10
bbbbabab
bbbabaaaaa

output

Copy

input

Copy

7 7
uiibwws
qhtkxcn

output

Copy

Note

For the first case:

abb from the first string and abab from the second string have LCS equal to abb.

The result is S(abb,abab)=(4⋅|abb|S(abb,abab)=(4⋅|abb|) - |abb||abb| - |abab||abab| = 4⋅3−3−4=54⋅3−3−4=5.

代码：

#include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=5e3+5;
const ll mod=1e9+7;
ll n,t,k,m,s,w,sum;
char x[maxn];
char y[maxn];
ll dp[maxn][maxn];
int main()
{
    cin>>n>>m;
    cin>>x;
    cin>>y;
    dp[0][0]=0;
    ll ans=0;
    for(int i=0;i<n;i++)
    {
        for(int j=0;j<m;j++)
        {
            if(x[i]==y[j])
            {
                dp[i+1][j+1]=max(dp[i][j],0ll)+2;
            }
            else
                dp[i+1][j+1]=max(dp[i+1][j],dp[i][j+1])-1;
            ans=max(ans,dp[i+1][j+1]);
        }

    }
    cout<<ans<<endl;
    return 0;
}