/*
JOI 2022 Problem 1
Solution: Sweep from left to right, sorting the queries and processing them in order.
The operations in the statement can be done using some basic math.
*/
#include <bits/stdc++.h>
using namespace std;
#define rep(i, a, b) for (int i = (a); i < (b); i++)
#define sz(x) ((int) x.size())
#define all(x) x.begin(), x.end()
#define debug if (true) cerr
using ll = long long;
using pii = pair<int, int>;
using vi = vector<int>;
void solve() {
int N; cin >> N;
vi A(N);
rep(i, 0, N) cin >> A[i];
int Q; cin >> Q;
vector<ll> X(Q);
rep(i, 0, Q) cin >> X[i];
ll piecesSoFar = 0;
int currQ = 0;
rep(i, 0, N) {
ll currSize = A[i];
ll currPieces = 1;
while (currSize % 2 == 0) {
currSize /= 2;
currPieces *= 2;
}
// Answer all queries in this range
piecesSoFar += currPieces;
while (currQ != Q && X[currQ] <= piecesSoFar) {
cout << currSize << '\n';
currQ++;
}
}
}
int main() {
ios::sync_with_stdio(0);
cin.tie(0);
cin.exceptions(cin.failbit);
solve();
return 0;
}
