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: Time Limit Exceeded)

#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <set>
#include <map>
#include <unordered_map>
#include <unordered_set>
#include <cstring>
#include <cmath>
#include <cstdlib>
#include <algorithm>
#include <random>
#include <iomanip>
#include <functional>
#include <cassert>

using namespace std;

typedef long long ll;

const int L = 1e5 + 7;
int dp[1 << 10][L];

int main() {
  ios_base::sync_with_stdio(false); cin.tie(0);
#ifdef LOCAL
  freopen("input.txt", "r", stdin);
#endif
  int n;
  cin >> n;
  vector <int> v(n), w(n);
  for (int i = 0; i < n; ++i) cin >> v[i] >> w[i];
  vector <int> h(n);
  vector <int> par(n);
  function <void(int)> dfs = [&] (int u) {
    if (h[u] <= 8) {
      if (u == 0) {
        dp[u][w[u]] = v[u];
      } else {
        for (int i = 0; i < L; ++i) {
          dp[u][i] = dp[par[u]][i];
          if (i >= w[u]) {
            dp[u][i] = max(dp[u][i], dp[par[u]][i - w[u]] + v[u]);
          }
        }
      }
      for (int i = 1; i < L; ++i) dp[u][i] = max(dp[u][i], dp[u][i - 1]);
    }

    if (2 * u + 1 < n) {
      h[2 * u + 1] = h[u] + 1;
      par[2 * u + 1] = u;
      dfs(2 * u + 1);
    }
    if (2 * u + 2 < n) {
      h[2 * u + 2] = h[u] + 1;
      par[2 * u + 2] = u;
      dfs(2 * u + 2);
    }
  };
  dfs(0);
  par[0] = -1;
  int q;
  cin >> q;
  for (int i = 0; i < q; ++i) {
    int u, l;
    cin >> u >> l;
    --u;
    if (h[u] <= 8) {
      cout << dp[u][l] << '\n';
    } else {
      vector <int> who;
      while (h[u] > 8) {
        who.push_back(u);
        u = par[u];
      }
      int ans = 0;
      for (int mask = 0; mask < (int)who.size(); ++i) {
        int sw = 0, sv = 0;
        for (int i = 0; i < (int)who.size(); ++i) {
          if (mask & (1 << i)) {
            sw += w[i];
            sv += v[i];
          }
        }
        if (sw <= l) {
          ans = max(ans, sv + dp[u][l - sw]);
        }
      }
      cout << ans << '\n';
    }   
  }

}