// header file
#include <bits/stdc++.h>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/tree_policy.hpp>
// pragma
#pragma GCC optimize("Ofast")
#pragma GCC optimize("unroll-loops")
// macros
#define endl "\n"
#define ll long long
#define mp make_pair
#define ins insert
#define lb lower_bound
#define pb push_back
#define ub upper_bound
#define lll __int128
#define fi first
#define se second
using namespace std;
using namespace __gnu_pbds;
typedef tree<int, null_type, less_equal<int>, rb_tree_tag, tree_order_statistics_node_update> ordered_multiset;
typedef tree<int, null_type, less<int>, rb_tree_tag,tree_order_statistics_node_update> ordered_set;
const int lim = 1 << 18;
bool used[lim];
struct segment_tree {
vector<int> a[2 * lim];
void query(int nd, int cl, int cr, int idx, vector<int> &v) {
// put in a and not in b
for(auto i : a[nd])
v.pb(i);
a[nd].clear();
if(cl == cr) {
return;
}
int mid = (cl + cr) / 2;
if(idx <= mid)
query(2 * nd, cl, mid, idx, v);
else
query(2 * nd + 1, mid + 1, cr, idx, v);
}
void update(int nd, int cl, int cr, int l, int r, int val) {
if(cl >= l && cr <= r) {
a[nd].pb(val);
}
else if(cl > r || cr < l) {
return;
}
else {
int mid = (cl + cr) / 2;
update(2 * nd, cl, mid, l, r, val);
update(2 * nd + 1, mid + 1, cr, l, r, val);
}
}
void erase(int nd, int cl, int cr, int l, int r, int val) {
if(cl >= l && cr <= r)
used[val] = 1;
else if(cl > r || cr < l)
return;
else {
int mid = (cl + cr) / 2;
erase(2 * nd, cl, mid, l, r, val);
erase(2 * nd + 1, mid + 1, cr, l, r, val);
}
}
};
int main() {
ios_base::sync_with_stdio(0); cin.tie(NULL);
// sub 1 -> greedy terjauh
// sub 2, 3 -> DP[L][R], bisa go left atau go right tiap iterasi, nanti max n^2 state, tiap transisi O(1)
// obs: mis skrg di range L, R
// belum tentu opt ambil furthest L/R
// bagi kasus jadi 3:
// extend L -> consider with minimum L
// extend R -> consider with maximum R
// extend L dan R -> technically should consider all?
// mst?
// possible case: a bisa ke b, b tdk bisa ke a (this makes every origin point unique)
// only consider 2 states, index of left and index of right
// only issue: extend BOTH L and R
// extend L/R -> can be done separately
// L dan R means taking a supersegment of the current segment
// dp memo simpan dr L, R ke 1, N itu butuh berapa
// obs: order does not really matter (only matter for unlock)
// jadi graph, jika a bisa ke b, maka edge a -> b
// minimum edges to connect containing nodes yg bs ke 1 dan node yg bs ke n
// "problem": common edges
// precalculate dr tiap node ke 1, n tanpa common edge
// nanti cri shortest dr sebuah node ke node lain
// pakai segment tree instead of pq, nanti tinggal min query buat cek
// bisa pakai vis/no vis trus di lazy prop di add param itu
// precalculate independent
// precalculate joint
// nanti mulai dari least cost dihitung dr 1, n nanti di invert edgenya
// pola:
// j j j j j (independent routing to 1 and to n)
int n;
cin >> n;
pair<int, int> a[n + 1];
for(int i = 1; i <= n; ++i)
cin >> a[i].fi >> a[i].se;
// buat segtree min query pair buat cari mana yg mau diubah
// buat dari 1
// fi -> cost
// se -> node
int from_one[n + 1], from_n[n + 1], total[n + 1];
memset(from_one, -1, sizeof(from_one));
memset(from_n, -1, sizeof(from_n));
memset(total, -1, sizeof(total));
priority_queue<pair<int, int>, vector<pair<int, int>>, greater<pair<int, int>>> pq;
segment_tree seg_one, seg_n, seg_all;
for(int i = 1; i <= n; ++i) {
if(a[i].fi == 1) {
pq.push(mp(0, i));
}
else {
seg_one.update(1, 0, lim - 1, a[i].fi, a[i].se, i);
}
}
while(pq.size()) {
int cost = pq.top().fi, nd = pq.top().se;
from_one[nd] = cost;
pq.pop();
vector<int> r;
seg_one.query(1, 0, lim - 1, nd, r);
for(auto i : r) {
if(used[i])
continue;
seg_one.erase(1, 0, lim - 1, a[i].fi, a[i].se, i);
pq.push(mp(cost + 1, i));
}
}
// buat dari n
for(int i = 1; i <= n; ++i) {
if(a[i].se == n) {
pq.push(mp(0, i));
}
else
seg_n.update(1, 0, lim - 1, a[i].fi, a[i].se, i);
}
memset(used, 0, sizeof(used));
while(pq.size()) {
int cost = pq.top().fi, nd = pq.top().se;
from_n[nd] = cost;
pq.pop();
vector<int> r;
seg_n.query(1, 0, lim - 1, nd, r);
for(auto i : r) {
if(used[i])
continue;
seg_n.erase(1, 0, lim - 1 ,a[i].fi, a[i].se, i);
pq.push(mp(cost + 1, i));
}
}
for(int i = 1; i <= n; ++i) {
if(from_one[i] != -1 && from_n[i] != -1) {
total[i] = from_one[i] + from_n[i] + 1;
pq.push(mp(from_one[i] + from_n[i], i));
seg_all.update(1, 0, lim - 1, a[i].fi, a[i].se, i);
}
}
bool vis[n + 1];
memset(vis, 0, sizeof(vis));
memset(used, 0, sizeof(used));
while(pq.size()) {
int cost = pq.top().fi, nd = pq.top().se;
pq.pop();
if(vis[nd])
continue;
total[nd] = min(total[nd], cost + 1);
vis[nd] = 1;
vector<int> r;
seg_all.erase(1, 0, lim - 1, a[nd].fi, a[nd].se, nd);
seg_all.query(1, 0, lim - 1, nd, r);
for(auto i : r) {
if(used[i])
continue;
pq.push(mp(cost + 1, i));
seg_all.erase(1, 0, lim - 1, a[i].fi, a[i].se, i);
}
}
//for(int i = 1; i <= n; ++i) {
// cout << from_one[i] << " " << from_n[i] << " " << total[i] << endl;
//}
// irrelevant as calculate for every n, jadi calc dulu bru input !!!
int q;
cin >> q;
for(int i = 0; i < q; ++i) {
int x;
cin >> x;
cout << total[x] << endl;
}
return 0;
}
