#pragma GCC optimize("O2") #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #define int ll #define INT128_MAX (__int128)(((unsigned __int128) 1 << ((sizeof(__int128) * __CHAR_BIT__) - 1)) - 1) #define INT128_MIN (-INT128_MAX - 1) #define clock chrono::steady_clock::now().time_since_epoch().count() #ifdef DEBUG #define dbg(x) cout << (#x) << " = " << (x) << '\n' #else #define dbg(x) #endif using namespace std; using ll = long long; using ull = unsigned long long; using ldb = long double; using pii = pair; using pll = pair; //#define double ldb template, class OP = plus> void pSum(rng &&v) { if (!v.empty()) for(T p = v[0]; T &x : v | views::drop(1)) x = p = OP()(p, x); } template, class OP> void pSum(rng &&v, OP op) { if (!v.empty()) for(T p = v[0]; T &x : v | views::drop(1)) x = p = op(p, x); } template T floorDiv(T a, T b) { if (b < 0) a *= -1, b *= -1; return a >= 0 ? a / b : (a - b + 1) / b; } template T ceilDiv(T a, T b) { if (b < 0) a *= -1, b *= -1; return a >= 0 ? (a + b - 1) / b : a / b; } template ostream& operator<<(ostream& os, const pair pr) { return os << pr.first << ' ' << pr.second; } template ostream& operator<<(ostream& os, const array &arr) { for(const T &X : arr) os << X << ' '; return os; } template ostream& operator<<(ostream& os, const vector &vec) { for(const T &X : vec) os << X << ' '; return os; } template ostream& operator<<(ostream& os, const set &s) { for(const T &x : s) os << x << ' '; return os; } signed main() { ios::sync_with_stdio(false), cin.tie(NULL); int n; cin >> n; vector> r(n); for(auto &[r1, r2] : r) cin >> r1 >> r2; ranges::sort(r); int ans = 0; priority_queue, greater> pq; for(int i = 0, j = 0; i < n; i = j) { while(j < n and r[i][0] == r[j][0]) j++; for(int k = i; k < j; k++) pq.push(r[k][1]); while(!pq.empty() and pq.top() <= r[i][0]) pq.pop(); ans = max(ans, (int)ssize(pq) - 1); } cout << ans << '\n'; return 0; }