#include #include #include #include #include #include #include #include #include #include #include #include #define rep(i,n) for(int i=0;i<(int)(n);++i) #define iter(a) __typeof(a.begin()) #define FOR(it,a) for(iter(a)it=a.begin();it!=a.end();++it) #define F first #define S second #define SZ(a) (int)((a).size()) #define sz(a) SZ(a) #define pb(a) push_back(a) #define mp(a,b) make_pair(a,b) #define ALL(a) (a).begin(),(a).end() using namespace std; typedef long long ll; typedef pair PI; typedef unsigned long long ull; #define PR(...) do{cerr << "line : " << __LINE__ << endl; pr(#__VA_ARGS__, __VA_ARGS__);}while(0); template void pr(const string& name, T t){ cerr << name << ": " << t << endl; } template void pr(const string& names, T t, Types ... rest) { auto comma_pos = names.find(','); cerr << names.substr(0, comma_pos) << ": " << t << ", "; auto next_name_pos = names.find_first_not_of(" \t\n", comma_pos + 1); pr(string(names, next_name_pos), rest ...); } template ostream& operator<< (ostream& o, const pair& v){return o << "(" << v.F << ", " << v.S << ")";} template ostream& operator<< (ostream& o, const vector& v){o << "{";rep(i,SZ(v)) o << (i?", ":"") << v[i];return o << "}";} const int dx[] = {0,1,0,-1}; const int dy[] = {-1,0,1,0}; #define endl '\n' int num[2100000]; bool npr[2100000]; int main(int argc, char *argv[]) { for(int i = 2; i<=2000000; ++i){ if(npr[i]) continue; for(int j = i; j <= 2000000; j+=i){ ++num[j]; npr[j] = 1; } } int n,k; cin >> n >> k; int ans = 0; rep(i,n+1) ans += num[i] >= k; cout << ans << endl; return 0; }