#include<bits/stdc++.h>
# define int long long
# define rep(i, j, k) for (int i = j; i <= k; ++i)
using namespace std;

int n, q, x[200005], lg2[200005], sta[20][200005], stb[20][200005];
void buildst()
{
    rep(i, 1, n) sta[0][i] = 2 * x[i] - x[i - 1], stb[0][i] = 2 * x[i] - x[i + 1];
    rep(p, 1, 18) rep(i, 1, n - (1 << p) + 1) sta[p][i] = max(sta[p - 1][i], sta[p - 1][i + (1 << (p - 1))]), stb[p][i] = min(stb[p - 1][i], stb[p - 1][i + (1 << (p - 1))]);
}

signed main()
{
//	freopen("ex.in", "r", stdin);
//	freopen("c.out", "w", stdout);
	ios::sync_with_stdio(false); cin.tie(0), cout.tie(0);
	cin >> n;
	rep(i, 1, n) cin >> x[i];
	buildst();
	cin >> q;
	rep(i, 1, q)
	{
	    int now, ans = 0; cin >> now;
	    if (now <= x[1]) {cout << x[n] - now << '\n'; continue;}
	    if (now >= x[n]) {cout << now - x[1] << '\n'; continue;}
	    int pos = upper_bound(x + 1, x + n + 1, now) - x - 1; //cerr << pos << endl;
	    if (now - x[pos] > x[pos + 1] - now) ++pos;
        ans = abs(now - x[pos]), now = x[pos];
        int l = pos - 1, r = pos + 1;
   //     cerr << l << ' ' << r << endl;
        while (l > 0 && r <= n)
        {
   //         cerr << l << ' ' << r << endl;
            if (now - x[l] <= x[r] - now)
            {
                if (l == 1) {ans += now - x[1], now = x[1], l = 0; continue;}
                int ppp = l;
                for (int i = 18; i >= 0; --i) if (ppp > (1 << i) + 1 && sta[i][ppp - (1 << i)] <= x[r]) ppp -= (1 << i);
                ans += now - x[ppp - 1], now = x[ppp - 1], l = ppp - 2;
     //           cerr << "gfdsgfds" << l << endl;
            }
            else
            {
                if (r == n) {ans += x[n] - now, now = x[n], r = n + 1; continue;}
                int ppp = r;
                for (int i = 18; i >= 0; --i) if (ppp + (1 << i) < n && stb[i][ppp] > x[l]) ppp += (1 << i);
                ans += x[ppp + 1] - now, r = ppp + 2, now = x[ppp + 1];
       //         cerr << "qwe" << r << endl;
            }
        }
        if (l > 0) ans += now - x[1]; else ans += x[n] - now;
        cout << ans << '\n';
	}
	return 0;
}