func_code_string: #include <bits/stdc++.h>   using namespace std;   template <typename T, typename U>   std::istream& operator>>(std::istream& i, pair<T, U>& p) {     i >> p.first >> p.second;     return i;   }   template <typename T>   std::istream& operator>>(std::istream& i, vector<T>& t) {     for (auto& v : t) {       i >> v;     }     return i;   }   template <typename T, typename U>   std::ostream& operator<<(std::ostream& o, const pair<T, U>& p) {     o << p.first <<     << p.second;     return o;   }   template <typename T>   std::ostream& operator<<(std::ostream& o, const vector<T>& t) {     if (t.empty()) o <<   n ;     for (size_t i = 0; i < t.size(); ++i) {       o << t[i] <<    n [i == t.size() - 1];     }     return o;   }   template <typename T>   using minheap = priority_queue<T, vector<T>, greater<T>>;   template <typename T>   using maxheap = priority_queue<T, vector<T>, less<T>>;   template <typename T>   bool in(T a, T b, T c) {     return a <= b && b < c;   }   unsigned int logceil(int first) {     return first ? 8 * sizeof(int) - __builtin_clz(first) : 0;   }   namespace std {   template <typename T, typename U>   struct hash<pair<T, U>> {     hash<T> t;     hash<U> u;     size_t operator()(const pair<T, U>& p) const {       return t(p.first) ^ (u(p.second) << 7);     }   };   }  // namespace std   template <typename T, typename F>   T bsh(T l, T h, const F& f) {     T r = -1, m;     while (l <= h) {       m = (l + h) / 2;       if (f(m)) {         l = m + 1;         r = m;       } else {         h = m - 1;       }     }     return r;   }   template <typename F>   double bshd(double l, double h, const F& f, double p = 1e-9) {     unsigned int r = 3 + (unsigned int)log2((h - l) / p);     while (r--) {       double m = (l + h) / 2;       if (f(m)) {         l = m;       } else {         h = m;       }     }     return (l + h) / 2;   }   template <typename T, typename F>   T bsl(T l, T h, const F& f) {     T r = -1, m;     while (l <= h) {       m = (l + h) / 2;       if (f(m)) {         h = m - 1;         r = m;       } else {         l = m + 1;       }     }     return r;   }   template <typename F>   double bsld(double l, double h, const F& f, double p = 1e-9) {     unsigned int r = 3 + (unsigned int)log2((h - l) / p);     while (r--) {       double m = (l + h) / 2;       if (f(m)) {         h = m;       } else {         l = m;       }     }     return (l + h) / 2;   }   template <typename T>   T gcd(T a, T b) {     if (a < b) swap(a, b);     return b ? gcd(b, a % b) : a;   }   template <typename T>   class vector2 : public vector<vector<T>> {    public:     vector2() {}     vector2(size_t a, size_t b, T t = T())         : vector<vector<T>>(a, vector<T>(b, t)) {}   };   template <typename T>   class vector3 : public vector<vector2<T>> {    public:     vector3() {}     vector3(size_t a, size_t b, size_t c, T t = T())         : vector<vector2<T>>(a, vector2<T>(b, c, t)) {}   };   template <typename T>   class vector4 : public vector<vector3<T>> {    public:     vector4() {}     vector4(size_t a, size_t b, size_t c, size_t d, T t = T())         : vector<vector3<T>>(a, vector3<T>(b, c, d, t)) {}   };   template <typename T>   class vector5 : public vector<vector4<T>> {    public:     vector5() {}     vector5(size_t a, size_t b, size_t c, size_t d, size_t e, T t = T())         : vector<vector4<T>>(a, vector4<T>(b, c, d, e, t)) {}   };   class TaskD {    public:     void solve(istream& cin, ostream& cout) {       long long N, L, R, K;       cin >> N >> L >> R >> K;       long long Q = (N + R - L + 1) % N;       if (Q == 0) Q = N;       if (K < Q) {         cout <<  -1 n ;         return;       }       long long ans = -1;       if (K <= 2 * Q) {         cout << min(N, K - Q + 1 + (N - Q)) << endl;         return;       }       for (long long rev = 1; rev <= 1000000; ++rev) {         for (long long take : {                  (K - Q + N) / (rev + 1),                  (K - Q + N) / (rev + 1),                  (K - Q + N) / (rev + 1),                  (2 * N + K - 2 * Q + 1) / (rev + 1),                  (2 * N + K - 2 * Q + 1) / (rev + 1),                  (2 * N + K - 2 * Q + 1) / (rev + 1),              }) {           long long s = take - N;           if (s < 0) continue;           if (s > N) continue;           long long left = (K - Q) - rev * take;           if (left < 0) continue;           if (left > Q) continue;           if (left > s) continue;           if (min(Q, left + 1) + (N - Q) < s) continue;           ans = max(ans, s);         }       }       for (long long take = 1; take <= 1000000; ++take) {         if (take < N) continue;         if (take > 2 * N) continue;         long long s = take - N;         for (long long rev = (K - 2 * Q) / take - 1; rev <= (K - Q) / take + 1;              ++rev) {           if (rev <= 0) continue;           long long left = (K - Q) - rev * take;           if (left < 0) continue;           if (left > Q) continue;           if (left > s) continue;           if (min(Q, left + 1) + (N - Q) < s) continue;           ans = max(ans, s);         }       }       cout << ans << endl;     }   };   int main() {     ios_base::sync_with_stdio(false);     cin.tie(nullptr);     cout.tie(nullptr);     TaskD solver;     std::istream& in(std::cin);     std::ostream& out(std::cout);     solver.solve(in, out);     return 0;   }   