#include <bits/stdc++.h>
using namespace std;
using ll = long long;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
// 1) Read input
int N;
cin >> N;
// (value, count) for each group
vector<pair<ll, ll>> pieces;
pieces.reserve(N); // To avoid multiple reallocations
// Process each A: factor out powers of 2
for(int i = 0; i < N; i++) {
ll A;
cin >> A;
ll count = 1;
while(A % 2 == 0) {
A /= 2;
count <<= 1; // same as count *= 2
}
pieces.emplace_back(A, count);
}
// 2) Build prefix sums: prefix_sum[i] = total pieces up to (and including) pieces[i-1]
// prefix_sum[0] = 0 by definition
vector<ll> prefix_sum(N + 1, 0LL);
for(int i = 1; i <= N; i++){
prefix_sum[i] = prefix_sum[i - 1] + pieces[i - 1].second;
}
// 3) Read queries
int Q;
cin >> Q;
// Problem states queries are sorted in non-decreasing order (X_j <= X_{j+1})
// so we can handle them in one pass
vector<ll> queries(Q);
for(int i = 0; i < Q; i++) {
cin >> queries[i];
}
// 4) Two-pointer approach
// i = index for prefix_sum (1..N)
// we move through queries (j from 0..Q-1) in sorted order
int i = 1; // Start comparing from prefix_sum[1]
for(int j = 0; j < Q; j++) {
ll x = queries[j];
// Move "i" forward until x <= prefix_sum[i]
// That means the x-th piece is in the group i-1
while(prefix_sum[i] < x) {
i++;
}
// Now we know x-th piece belongs to pieces[i - 1]
cout << pieces[i - 1].first << "\n";
}
return 0;
}
