#define _USE_MATH_DEFINES #include using namespace std; #define FOR(i,m,n) for(int i=(m);i<(n);++i) #define REP(i,n) FOR(i,0,n) #define ALL(v) (v).begin(),(v).end() using ll = long long; constexpr int INF = 0x3f3f3f3f; constexpr long long LINF = 0x3f3f3f3f3f3f3f3fLL; constexpr double EPS = 1e-8; constexpr int MOD = 1000000007; // constexpr int MOD = 998244353; constexpr int DY4[]{1, 0, -1, 0}, DX4[]{0, -1, 0, 1}; constexpr int DY8[]{1, 1, 0, -1, -1, -1, 0, 1}; constexpr int DX8[]{0, -1, -1, -1, 0, 1, 1, 1}; template inline bool chmax(T& a, U b) { return a < b ? (a = b, true) : false; } template inline bool chmin(T& a, U b) { return a > b ? (a = b, true) : false; } struct IOSetup { IOSetup() { std::cin.tie(nullptr); std::ios_base::sync_with_stdio(false); std::cout << fixed << setprecision(20); } } iosetup; std::vector prime_sieve(const int n, const bool get_only_prime) { std::vector smallest_prime_factor(n + 1), prime; std::iota(smallest_prime_factor.begin(), smallest_prime_factor.end(), 0); for (int i = 2; i <= n; ++i) { if (smallest_prime_factor[i] == i) prime.emplace_back(i); for (const int p : prime) { if (i * p > n || p > smallest_prime_factor[i]) break; smallest_prime_factor[i * p] = p; } } return get_only_prime ? prime : smallest_prime_factor; } struct Divisor { const std::vector smallest_prime_factor; explicit Divisor(const int n) : smallest_prime_factor(prime_sieve(n, false)) {} std::vector query(int n) const { std::vector res{1}; while (n > 1) { const int prime_factor = smallest_prime_factor[n], d = res.size(); int tmp = 1; for (; n % prime_factor == 0; n /= prime_factor) { tmp *= prime_factor; for (int i = 0; i < d; ++i) { res.emplace_back(res[i] * tmp); } } } // std::sort(res.begin(), res.end()); return res; } }; int main() { int n, m; cin >> n >> m; vector is_open(n + 1, false); while (m--) { int a; cin >> a; is_open[a] = true; } Divisor divisor(n); vector b(n + 1, 0); for (int i = 1; i <= n; ++i) { for (int d : divisor.query(i)) ++b[d]; } int ans = 0; for (int i = n; i >= 1; --i) { if (is_open[i] != b[i] % 2) { ++ans; for (int d : divisor.query(i)) --b[d]; } } cout << ans << '\n'; return 0; }