//yukicoder@cpp17 #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include #include using namespace std; using ll = long long; using P = pair; const ll MOD = 998244353; const ll MODx = 1000000007; const int INF = (1<<30)-1; const ll LINF = (1LL<<62LL)-1; const double EPS = (1e-10); P ar4[4] = {{0,1},{0,-1},{1,0},{-1,0}}; P ar8[8] = {{-1,-1},{-1,0},{-1,1},{0,-1},{0,1},{1,-1},{1,0},{1,1}}; template vector make_vector(size_t a, T b) { return vector(a, b); } template auto make_vector(size_t a, Ts... ts) { return vector(a, make_vector(ts...)); } /* 確認ポイント cout << fixed << setprecision(n) << 小数計算//n桁の小数表記になる 計算量は変わらないが楽できるシリーズ min(max)_element(iter,iter)で一番小さい(大きい)値のポインタが帰ってくる count(iter,iter,int)でintがiterからiterの間にいくつあったかを取得できる */ /* comment outed because can cause bugs __attribute__((constructor)) void initial() { cin.tie(0); ios::sync_with_stdio(false); } */ struct d{ map g; d (long long n){ add(n); } void add(long long n){ for(int i = 2; n >= i*i; i++){ if(n%i == 0){ int nw = 0; while(n%i == 0){ n/=i; nw++; } g[i] += nw; } } if(n != 1)g[n] += 1; } bool operator<(const d z)const{ vector> A1,A2; for(auto e: z.g){ A1.push_back(e); } for(auto e: g){ A2.push_back(e); } for(int i = 0; min(A1.size(), A2.size()) > i; i++){ if(A1[i].first < A2[i].first)return true; else if(A1[i].first > A2[i].first)return false; else if(A1[i].second < A2[i].second)return true; else if(A1[i].second > A2[i].second)return false; } if(A1.size() < A2.size())return true; return false; } bool operator==(const d z)const{ vector> A1,A2; for(auto e: z.g){ A1.push_back(e); } for(auto e: g){ A2.push_back(e); } if(A1.size() != A2.size())return false; for(int i = 0; A2.size() > i; i++){ if(A1[i].first != A2[i].first || A1[i].second != A2[i].second)return false; } return true; } friend ostream &operator<<(ostream &os, const d z){ for(auto g: z.g){ os << " (" << g.first << ", " << g.second << ") "; } return os; } }; int main(){ int n,k;cin>>n>>k; vector A(n); for(int i = 0; n > i; i++)cin>>A[i]; set Z; for(int i = 0; (1< i; i++){ if(__builtin_popcount(i) < k)continue; long long x = 0; d a1(1); for(int j = 0; n > j; j++){ if(i & (1<