/** * author: otera **/ #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; //#define int long long typedef long long ll; typedef unsigned long long ul; typedef unsigned int ui; typedef long double ld; const int inf=1e9+7; const ll INF=1LL<<60 ; const ll mod=1e9+7 ; #define rep(i,n) for(int i=0;i=0;i--) #define Rep(i,sta,n) for(int i=sta;i=1;i--) #define Rep1(i,sta,n) for(int i=sta;i<=n;i++) typedef complex Point; const ld eps = 1e-8; const ld pi = acos(-1.0); typedef pair P; typedef pair LDP; typedef pair LP; #define fr first #define sc second #define all(c) c.begin(),c.end() #define pb push_back #define debug(x) cerr << #x << " = " << (x) << endl; template inline bool chmax(T& a, T b) { if (a < b) { a = b; return 1; } return 0; } template inline bool chmin(T& a, T b) { if (a > b) { a = b; return 1; } return 0; } // ローリングハッシュ // 二分探索で LCP を求める機能つき struct RollingHash { static const int base1 = 1007, base2 = 2009; static const int mod1 = 1000000007, mod2 = 1000000009; vector hash1, hash2, power1, power2; // construct RollingHash(const string &S) { int n = (int)S.size(); hash1.assign(n+1, 0); hash2.assign(n+1, 0); power1.assign(n+1, 1); power2.assign(n+1, 1); for (int i = 0; i < n; ++i) { hash1[i+1] = (hash1[i] * base1 + S[i]) % mod1; hash2[i+1] = (hash2[i] * base2 + S[i]) % mod2; power1[i+1] = (power1[i] * base1) % mod1; power2[i+1] = (power2[i] * base2) % mod2; } } // get hash of S[left:right) inline pair get(int l, int r) const { long long res1 = hash1[r] - hash1[l] * power1[r-l] % mod1; if (res1 < 0) res1 += mod1; long long res2 = hash2[r] - hash2[l] * power2[r-l] % mod2; if (res2 < 0) res2 += mod2; return {res1, res2}; } // get lcp of S[a:] and S[b:] inline int getLCP(int a, int b) const { int len = min((int)hash1.size()-a, (int)hash1.size()-b); int low = 0, high = len; while (high - low > 1) { int mid = (low + high) >> 1; if (get(a, a+mid) != get(b, b+mid)) high = mid; else low = mid; } return low; } // get lcp of S[a:] and T[b:] inline int getLCP(const RollingHash &T, int a, int b) const { int len = min((int)hash1.size()-a, (int)hash1.size()-b); int low = 0, high = len; while (high - low > 1) { int mid = (low + high) >> 1; if (get(a, a+mid) != T.get(b, b+mid)) high = mid; else low = mid; } return low; } }; void solve() { string s, t; cin >> s >> t; int n = (int)s.size(), m = (int)t.size(); RollingHash rhs(s), rht(t); int i = 0, ans = 0; for(; i < n - m;) { if(rhs.get(i, i + m) == rht.get(0, m)) { if(m == 1) { cout << -1 << endl; return; } ++ ans; i += m - 1; } ++ i; } cout << ans << endl; } signed main() { ios::sync_with_stdio(false); cin.tie(0); //cout << fixed << setprecision(10); //int t; cin >> t; rep(i, t)solve(); solve(); return 0; }