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: Accepted)

#include <bits/stdc++.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define all(x) (x).begin(),(x).end()
using namespace std;
using ll = long long;
using P = pair<int,int>;
template <class T> void chmin(T &a, const T &b) noexcept { if (b < a) a = b; }
template <class T> void chmax(T &a, const T &b) noexcept { if (a < b) a = b; }
void debug_out() { cout << "\n"; }
template <class T, class... Args>
void debug_out(const T &x, const Args &... args) { cout << x << " "; debug_out(args...);}
#ifdef _DEBUG
  #define debug(...) debug_out(__VA_ARGS__)
#else
  #define debug(...) 
#endif

struct item_t {
  ll w, v;
  item_t(ll w, ll v):w(w),v(v){}
  item_t():w(0),v(0){}
  void operator+=(const item_t &r) {
    w += r.w;
    v += r.v;
  }
  bool operator<(const item_t &r) const { return w < r.w; }
};

vector<item_t> rekkyo(vector<item_t> item) {
  vector<item_t> a(1, item_t(0,0));
  for (auto i : item) {
    vector<item_t> b = a, c;
    for (auto &e : b) {
      e += i;
    }
    std::merge(a.begin(), a.end(), b.begin(), b.end(), std::back_inserter(c));
    a = c;
  }
  return a;
}

int main() {
  std::cin.tie(nullptr);
  std::ios_base::sync_with_stdio(false);
  std::cout << std::fixed << std::setprecision(15);
  int n;
  cin >> n;
  vector<item_t> item(n+1); //1-indexed
  rep(i,n) cin >> item[i+1].v >> item[i+1].w;
  int q;
  cin >> q;
  while(q--) {
    int v, L;
    cin >> v >> L;
    vector<item_t> A;
    while(v > 0) {  
      A.emplace_back(item[v]);
      v /= 2;
    }
    int m = A.size();
    auto a = rekkyo({A.begin(), A.begin() + m/2});
    auto b = rekkyo({A.begin() + m/2, A.end()});
    ll maxv = 0;
    for (auto&& x : b) { //x.v <- x.w以下で達成できる価値の最大値
      chmax(maxv, x.v);
      x.v = maxv;
    }
    reverse(all(a));
    ll ans = 0;
    int r = 0;
    for (auto&& x : a) {
      while(r+1 < (int)b.size() && x.w + b[r+1].w <= L) {
        r++;
      }
      if (x.w + b[r].w > L) continue;  
      chmax(ans, x.v + b[r].v);
    }
    cout << ans << "\n";
  }
  return 0;
}