#pragma GCC optimization ("O3") #include using namespace std; using ll = int; using vec = vector; using mat = vector; using pll = pair; using dvec = vector; using dmat = vector; #define INF (1LL<<61) #define MOD 1000000007LL //#define MOD 998244353LL #define PR(x) cout << (x) << endl #define PS(x) cout << (x) << " " #define REP(i,m,n) for(ll (i)=(m),(i_len)=(n);(i)<(i_len);++(i)) #define FORE(i,v) for(auto (i):v) #define ALL(x) (x).begin(), (x).end() #define SZ(x) ((ll)(x).size()) #define REV(x) reverse(ALL((x))) #define ASC(x) sort(ALL((x))) #define DESC(x) {ASC((x)); REV((x));} #define BIT(s,i) (((s)>>(i))&1) #define pb push_back #define fi first #define se second template inline int chmin(T& a, T b) {if(a>b) {a=b; return 1;} return 0;} template inline int chmax(T& a, T b) {if(a=MOD) x-=MOD; return *this;} mint& operator-=(const mint& a) {if((x+=MOD-a.x)>=MOD) x-=MOD; return *this;} mint& operator*=(const mint& a) {(x*=a.x)%=MOD; return *this;} mint operator+(const mint& a) const {mint b(*this); return b+=a;} mint operator-(const mint& a) const {mint b(*this); return b-=a;} mint operator*(const mint& a) const {mint b(*this); return b*=a;} mint pow(ll t) const {if(!t) return 1; mint a=pow(t>>1); return (t&1?*this*a:a)*a;} mint inv() const {return pow(MOD-2);} mint& operator/=(const mint& a) {return *this*=a.inv();} mint operator/(const mint& a) const {mint b(*this); return b/=a;} }; istream &operator>>(istream& is, mint& a) {ll t; is>>t; a=t; return is;} ostream &operator<<(ostream& os, const mint& a) {return os<; using mmat = vector; map factorize(ll n) { map p; ll t = n; for(ll i=2; i*i<=t; ++i) { ll cnt = 0; if(t%i) continue; while(t%i == 0) { t /= i; ++cnt; } p[i] = cnt; } if(t != 1) p[t] = 1; return p; } int main() { ll N; cin >> N; vec A(N+1); REP(i,1,N+1) cin >> A[i]; vector> C(N+1); set S; REP(i,1,N+1) { C[i] = factorize(A[i]); FORE(p,C[i]) S.insert(p.fi); } REP(i,1,N+1) { FORE(p,S) C[i][p] += C[i-1][p]; } ll Q; cin >> Q; while(Q--) { ll P, L, R; cin >> P >> L >> R; bool ans = true; map F = factorize(P); FORE(p,F) S.insert(p.fi); FORE(p,S) { if(F[p] > C[R][p]-C[L-1][p]) { ans = false; break; } } if(ans) PR("Yes"); else PR("NO"); } return 0; } /* */