#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::{Write, BufWriter};
// https://qiita.com/tanakh/items/0ba42c7ca36cd29d0ac8
macro_rules! input {
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes.by_ref().map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};
    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };
    ($next:expr, $t:ty) => ($next().parse::<$t>().expect("Parse error"));
}

/**
 * Segment Tree. This data structure is useful for fast folding on intervals of an array
 * whose elements are elements of monoid I. Note that constructing this tree requires the identity
 * element of I and the operation of I.
 * Verified by: yukicoder No. 259 (http://yukicoder.me/submissions/100581)
 *              AGC015-E (http://agc015.contest.atcoder.jp/submissions/1461001)
 */
struct SegTree<I, BiOp> {
    n: usize,
    dat: Vec<I>,
    op: BiOp,
    e: I,
}

impl<I, BiOp> SegTree<I, BiOp>
    where BiOp: Fn(I, I) -> I,
          I: Copy {
    pub fn new(n_: usize, op: BiOp, e: I) -> Self {
        let mut n = 1;
        while n < n_ { n *= 2; } // n is a power of 2
        SegTree {n: n, dat: vec![e; 2 * n - 1], op: op, e: e}
    }
    /* ary[k] <- v */
    pub fn update(&mut self, idx: usize, v: I) {
        let mut k = idx + self.n - 1;
        self.dat[k] = v;
        while k > 0 {
            k = (k - 1) / 2;
            self.dat[k] = (self.op)(self.dat[2 * k + 1], self.dat[2 * k + 2]);
        }
    }
    /* [a, b) (note: half-inclusive)
     * http://proc-cpuinfo.fixstars.com/2017/07/optimize-segment-tree/ */
    pub fn query(&self, mut a: usize, mut b: usize) -> I {
        let mut left = self.e;
        let mut right = self.e;
        a += self.n - 1;
        b += self.n - 1;
        while a < b {
            if (a & 1) == 0 {
                left = (self.op)(left, self.dat[a]);
            }
            if (b & 1) == 0 {
                right = (self.op)(self.dat[b - 1], right);
            }
            a = a / 2;
            b = (b - 1) / 2;
        }
        (self.op)(left, right)
    }
}

fn solve() {
    let out = std::io::stdout();
    let mut out = BufWriter::new(out.lock());
    macro_rules! puts {
        ($($format:tt)*) => (let _ = write!(out,$($format)*););
    }
    input! {
        n: usize,
        a: [i64; n],
        q: usize,
        b: [i64; q],
    }
    let mut st = SegTree::new(n, |x, y| x | y, 0i64);
    for i in 0..n {
        st.update(i, a[i]);
    }
    let whole = (1i64 << 60) - 1;
    for b in b {
        let mut targ = b;
        let mut rem = whole;
        let mut cur = n;
        let mut ok = true;
        let mut ans = 0;
        let last = a[n - 1];
        if (last & b) != last {
            targ ^= rem;
            ans += 1;
            if (targ & last) != last {
                ok = false;
            }
        }
        while cur > 0 && rem != 0 && ok {
            assert_eq!(rem & targ, targ);
            let mut fail = 0;
            let mut pass = cur + 1;
            while pass - fail > 1 {
                let mid = (pass + fail) / 2;
                let o = st.query(mid - 1, cur);
                if ((o & rem) | targ) == targ {
                    pass = mid;
                } else {
                    fail = mid;
                }
            }
            if pass == cur + 1 {
                ok = false;
                break;
            }
            let o = st.query(pass - 1, cur);
            // eprintln!("b = {}, [{}, {}) o = {} ({} <= {})", b, pass - 1, cur, o, targ, rem);
            rem &= !o;
            targ &= rem;
            cur = pass - 1;
            if cur != 0 {
                targ ^= rem;
                ans += 1;
            }
        }
        // TODO: prove it's ok to flip "before the first element"
        ok &= targ == 0 || targ == rem;
        if targ == rem {
            ans += 1;
        }
        puts!("{}\n", if ok { ans } else { -1 });
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}