Problem:
Score : 700 points

Problem Statement

We have a rooted binary tree with N vertices, where the vertices are numbered 1 to N.
Vertex 1 is the root, and the parent of Vertex i (i \geq 2) is Vertex \left[ \frac{i}{2} \right].

Each vertex has one item in it. The item in Vertex i has a value of V_i and a weight of W_i.
Now, process the following query Q times:

Given are a vertex v of the tree and a positive integer L.
Let us choose some (possibly none) of the items in v and the ancestors of v so that their total weight is at most L.
Find the maximum possible total value of the chosen items.

Here, Vertex u is said to be an ancestor of Vertex v when u is an indirect parent of v, that is, there exists a sequence of vertices w_1,w_2,\ldots,w_k (k\geq 2) where w_1=v, w_k=u, and w_{i+1} is the parent of w_i for each i.

Constraints

All values in input are integers.

1 \leq N < 2^{18}

1 \leq Q \leq 10^5

1 \leq V_i \leq 10^5

1 \leq W_i \leq 10^5

For the values v and L given in each query, 1 \leq v \leq N and 1 \leq L \leq 10^5.

Input

Let v_i and L_i be the values v and L given in the i-th query.
Then, Input is given from Standard Input in the following format:

N
V_1 W_1
:
V_N W_N
Q
v_1 L_1
:
v_Q L_Q

Output

For each integer i from 1 through Q,
the i-th line should contain the response to the i-th query.

Sample Input 1

3
1 2
2 3
3 4
3
1 1
2 5
3 5

Sample Output 1

0
3
3

In the first query, we are given only one choice: the item with (V, W)=(1,2). Since L = 1, we cannot actually choose it, so our response should be 0.

In the second query, we are given two choices: the items with (V, W)=(1,2) and (V, W)=(2,3). Since L = 5, we can choose both of them, so our response should be 3.

Sample Input 2

15
123 119
129 120
132 112
126 109
118 103
115 109
102 100
130 120
105 105
132 115
104 102
107 107
127 116
121 104
121 115
8
8 234
9 244
10 226
11 227
12 240
13 237
14 206
15 227

Sample Output 2

256
255
250
247
255
259
223
253

Solution in C++ (GCC 9.2.1) (judge verdict: Runtime Error)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
using namespace std;
#define FOR(i,m,n) for(int i=(m);i<(n);++i)
#define REP(i,n) FOR(i,0,n)
#define ALL(v) (v).begin(),(v).end()
using ll = long long;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3fLL;
const double EPS = 1e-8;
const int MOD = 1000000007;
// const int MOD = 998244353;
const int dy[] = {1, 0, -1, 0}, dx[] = {0, -1, 0, 1};
const int dy8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dx8[] = {0, -1, -1, -1, 0, 1, 1, 1};
template <typename T, typename U> inline bool chmax(T &a, U b) { return a < b ? (a = b, true) : false; }
template <typename T, typename U> inline bool chmin(T &a, U b) { return a > b ? (a = b, true) : false; }
struct IOSetup {
  IOSetup() {
    cin.tie(nullptr);
    ios_base::sync_with_stdio(false);
    cout << fixed << setprecision(20);
  }
} iosetup;

template <typename Monoid>
struct RMQ {
  RMQ(int sz, const Monoid UNITY) : UNITY(UNITY) {
    init(sz);
    dat.assign((n << 1) - 1, UNITY);
  }

  RMQ(const vector<Monoid> &a, const Monoid UNITY) : UNITY(UNITY) {
    int a_sz = a.size();
    init(a_sz);
    dat.resize((n << 1) - 1);
    REP(i, a_sz) dat[n - 1 + i] = a[i];
    for (int i = n - 2; i >= 0; --i) dat[i] = max(dat[(i << 1) + 1], dat[(i << 1) + 2]);
  }

  void update(int node, Monoid val) {
    node += n - 1;
    dat[node] = val;
    while (node > 0) {
      node = (node - 1) >> 1;
      dat[node] = max(dat[(node << 1) + 1], dat[(node << 1) + 2]);
    }
  }

  Monoid query(int a, int b) { return query(a, b, 0, 0, n); }

  int find(int a, int b, Monoid val) { return find(a, b, val, 0, 0, n); }

  Monoid operator[](const int idx) const { return dat[idx + n - 1]; }

private:
  int n = 1;
  const Monoid UNITY;
  vector<Monoid> dat;

  void init(int sz) { while (n < sz) n <<= 1; }

  Monoid query(int a, int b, int node, int left, int right) {
    if (right <= a || b <= left) return UNITY;
    if (a <= left && right <= b) return dat[node];
    return max(query(a, b, (node << 1) + 1, left, (left + right) >> 1), query(a, b, (node << 1) + 2, (left + right) >> 1, right));
  }

  int find(int a, int b, Monoid val, int node, int left, int right) {
    if (dat[node] < val || right <= a || b <= left) return -1;
    if (right - left == 1) return node - (n - 1);
    int res_l = find(a, b, val, (node << 1) + 1, left, (left + right) >> 1);
    if (res_l != -1) return res_l;
    return find(a, b, val, (node << 1) + 2, (left + right) >> 1, right);
  }
};

int main() {
  int n; cin >> n;
  vector<int> v(n), w(n); REP(i, n) cin >> v[i] >> w[i];
  int q; cin >> q;
  vector<vector<pair<int, pair<int, int>>>> qs(n);
  REP(i, q) {
    int vi, l; cin >> vi >> l; --vi;
    vector<pair<int, int>> now{{0, l}};
    while (vi >> 9) {
      vector<pair<int, int>> nx;
      for (auto [a, b] : now) {
        nx.emplace_back(a, b);
        if (b - w[vi] >= 0) nx.emplace_back(a + v[vi], b - w[vi]);
      }
      now.swap(nx);
      vi >>= 1;
    }
    for (auto &pr : now) qs[vi].emplace_back(i, pr);
  }
  vector<int> ans(q, 0);
  const int X = 100000;
  RMQ<int> rmq(X + 1, -INF);
  rmq.update(0, 0);
  set<int> st{0};
  function<void(int)> rec = [&](int ver) {
    if (ver >> 10) return;
    for (auto pr : qs[ver]) {
      int idx = pr.first;
      auto [vv, lda] = pr.second;
      chmax(ans[idx], rmq.query(0, lda + 1) + vv);
      if (lda >= w[ver]) chmax(ans[idx], rmq.query(0, lda - w[ver] + 1) + v[ver] + vv);
    }
    if (ver * 2 + 1 >= n) return;
    vector<pair<int, int>> chan;
    set<int> nx;
    for (int e : st) {
      nx.emplace(e);
      int now = -e;
      if (now + w[ver] <= X && rmq[now] + v[ver] > rmq[now + w[ver]]) {
        chan.emplace_back(now + w[ver], rmq[now + w[ver]]);
        rmq.update(now + w[ver], rmq[now] + v[ver]);
        nx.emplace(-(now + w[ver]));
      }
    }
    swap(st, nx);
    rec(ver * 2 + 1);
    rec(ver * 2 + 2);
    swap(st, nx);
    for (auto [a, b] : chan) rmq.update(a, b);
  };
  rec(0);
  REP(i, q) cout << ans[i] << '\n';
  return 0;
}