/*
* Author: Nonoze
* Created: Saturday 27/12/2025
*/
#include <bits/stdc++.h>
using namespace std;
#ifndef DEBUG
#define dbg(...)
#endif
// #define cout cerr << "OUT: "
#define endl '\n'
#define endlfl '\n' << flush
#define quit(x) return (void)(cout << x << endl)
template<typename T> void read(T& x) { cin >> x; }
template<typename T1, typename T2> void read(pair<T1, T2>& p) { read(p.first), read(p.second); }
template<typename T> void read(vector<T>& v) { for (auto& x : v) read(x); }
template<typename T1, typename T2> void read(T1& x, T2& y) { read(x), read(y); }
template<typename T1, typename T2, typename T3> void read(T1& x, T2& y, T3& z) { read(x), read(y), read(z); }
template<typename T1, typename T2, typename T3, typename T4> void read(T1& x, T2& y, T3& z, T4& zz) { read(x), read(y), read(z), read(zz); }
template<typename T> void print(vector<T>& v) { for (auto& x : v) cout << x << ' '; cout << endl; }
#define sz(x) (int)(x.size())
#define all(x) (x).begin(), (x).end()
#define rall(x) (x).rbegin(), (x).rend()
#define make_unique(v) sort(all(v)), v.erase(unique(all(v)), (v).end())
#define pb push_back
#define mp(a, b) make_pair(a, b)
#define fi first
#define se second
#define cmin(a, b) a = min(a, b)
#define cmax(a, b) a = max(a, b)
#define YES cout << "YES" << endl
#define NO cout << "NO" << endl
#define QYES quit("YES")
#define QNO quit("NO")
#define int long long
#define double long double
const int inf = numeric_limits<int>::max() / 4;
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
const int MOD = 1e9+7, LOG=20;
void solve();
signed main() {
ios::sync_with_stdio(0);
cin.tie(0);
int tt=1;
// cin >> tt;
while(tt--) solve();
return 0;
}
int n, k, m, q;
string s, t;
vector<vector<int>> memo1, memo2;
int dpright(int i, int j) {
if (j==m || i==2*n) return 0;
if (memo1[i][j]!=-1) return memo1[i][j];
if (s[i]==t[j]) return memo1[i][j] = 1 + dpright(i+1, j+1);
return memo1[i][j] = max(dpright(i+1, j), dpright(i, j+1));
}
int dpleft(int i, int j) {
if (j<0 || i<0) return 0;
if (memo2[i][j]!=-1) return memo2[i][j];
if (s[i]==t[j]) return memo2[i][j] = 1 + dpleft(i-1, j-1);
return memo2[i][j] = max(dpleft(i-1, j), dpleft(i, j-1));
}
void solve() {
read(s, t); if (sz(s)>sz(t)) swap(s, t);
n = sz(s), m = sz(t); s=s+s;
memo1.assign(2*n+1, vector<int>(m+1, -1)), memo2.assign(2*n+1, vector<int>(m, -1));
int ans=0;
for (int i=0; i<=n; i++) {
for (int j=0; j<=m; j++) {
cmax(ans, dpleft(n-1, j-1)+dpright(n, j));
}
}
cout << ans << endl;
}
