#include <bits/stdc++.h>
using namespace std;
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx,tune=native")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("Ofast")
#pragma GCC optimize("O3")
#pragma GCC target("avx")
const long long INF = numeric_limits<long long>::max();
const long double PI = 3.1415926535898;
const long long MOD = 998244353;
const long long LIM = 500005;
long long fpow(long long x, long long y) {
  long long temp;
  if (y == 0) return 1;
  temp = fpow(x, y / 2);
  if (y % 2 == 0)
    return (temp * temp) % MOD;
  else
    return (x * ((temp * temp) % MOD)) % MOD;
}
long long gcd(long long a, long long b) {
  if (b == 0) return a;
  return gcd(b, a % b);
}
bool isPrime(long long n) {
  if (n <= 1) return false;
  if (n <= 3) return true;
  if (n % 2 == 0 || n % 3 == 0) return false;
  for (long long i = 5; i * i <= n; i = i + 6)
    if (n % i == 0 || n % (i + 2) == 0) return false;
  return true;
}
long long modInverse(long long n, long long p) { return fpow(n, p - 2); }
map<long long, long long> primeFactors(long long n) {
  map<long long, long long> mp;
  while (n % 2 == 0) {
    if (mp.count(2) == 1)
      mp[2]++;
    else
      mp.insert({2, 1});
    n = n / 2;
  }
  for (long long i = 3; i <= sqrt(n); i = i + 2) {
    while (n % i == 0) {
      if (mp.count(i) == 1)
        mp[i]++;
      else
        mp.insert({i, 1});
      n = n / i;
    }
  }
  if (n > 2) {
    if (mp.count(n) == 1)
      mp[n]++;
    else
      mp.insert({n, 1});
  }
  return mp;
}
int32_t main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n, m;
  cin >> n >> m;
  string a, b;
  cin >> a >> b;
  long long dp[n + 1][m + 1];
  for (long long i = 0; i <= n; i++) {
    for (long long j = 0; j <= m; j++) dp[i][j] = 0;
  }
  for (long long i = 1; i <= n; i++) {
    for (long long j = 1; j <= m; j++) {
      if (a[i - 1] == b[j - 1])
        dp[i][j] = 2 + dp[i - 1][j - 1];
      else
        dp[i][j] = max(dp[i - 1][j], dp[i][j - 1]) - 1;
      if (dp[i][j] < 0) dp[i][j] = 0;
    }
  }
  long long ans = 0;
  for (long long i = 1; i <= n; i++)
    for (long long j = 1; j <= m; j++) ans = max(ans, dp[i][j]);
  cout << ans << "\n";
  return 0;
}
