#include using namespace std; using T = int; struct XAB{ int x; int a; int b; XAB(int _x=0, int _a=0, int _b=0):x(_x), a(_a), b(_b){} }; struct BIT{ vector bit; BIT(int n) : bit(n, 0){} T sum(int j) const { assert(j < (int)bit.size()); T s = 0; for (; j >= 0; j = (j & (j + 1)) - 1) s += bit[j]; return s; } T total() const { return sum(bit.size() - 1);} T count_larger(int k) const { return total() - sum(k - 1);} void add(int k, T a) {for (; k < bit.size(); k |= k + 1) bit[k] += a;} }; vector ignore_3complete(const vector & xabs){ vector result; for (auto xab : xabs){ if (xab.x != 3) result.emplace_back(xab); } return result; } int count_2complete(const vector & xabs){ int c = 0; for (auto &xab : xabs) if (xab.x == 2) ++c; return c; } int count3(const vector & B, int M){ int solved0 = B[0].total(); int solved1 = B[1].total(); int solved2 = B[2].total(); int solved3 = B[3].total(); int target = M - solved2 - solved3; if (target <= 0) return solved3; if (target > solved1) return 1e9; int OK = 0; int NG = 100001; while (NG > (OK + 1)){ int mid = (OK + NG)/2; if (B[1].count_larger(mid) >= target){ OK = mid; }else{ NG = mid; } } return solved3 + B[2].count_larger(OK); } void update_B(const vector>> & data, vector & B, int SA){ for (int i = 2; i >= 0; --i){ for (auto b : data[i][SA]){ B[i + 1].add(b, -1); B[i].add(b, 1); } } } int solve(vector & xabs, int M){ if (M <= 0) return 0; int n_2complete = count_2complete(xabs); if (n_2complete >= M) return 0; vector>> data(3, vector>(100005)); vector B(4, BIT(100002)); for (auto &xab : xabs){ data[xab.x][xab.a].push_back(xab.b); B[xab.x + 1].add(xab.b, 1); } int record = 1e9; for (int SA = 0; SA <= 100001; ++SA){ int score = count3(B, M); record = min(record, score); update_B(data, B, SA); } return record; } int main() { cin.tie(0); ios::sync_with_stdio(false); int N, M; cin >> N >> M; vector xabs(N); for (auto & xab : xabs) cin >> xab.x >> xab.a >> xab.b; auto xabs012 = ignore_3complete(xabs); int n_3complete = N - xabs012.size(); cout << solve(xabs012, M - n_3complete) + n_3complete << endl; return 0; }