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>
using namespace std;

typedef long long ll;

ll N;
vector<ll> V, W;
ll Q;
const ll K = 9, L = 100000;
ll I;
vector<vector<ll>> Dp;
vector<pair<ll, ll>> C;

ll solve(ll i, ll l) {
  C.clear();
  C.push_back({l, 0});
  while (i > I) {
    ll n = C.size();
    for (ll k = 0; k < n; ++k) {
      ll l, v;
      tie(l, v) = C[k];
      if (l >= W[i]) C.push_back({l - W[i], v + V[i]});
    }
    i /= 2;
  }
  ll res = 0;
  for (auto lv : C) {
    ll l, v;
    tie(l, v) = lv;
    res = max(res, v + Dp[i][l]);
  }
  return res;
}
int main() {
  cin >> N;
  V.resize(N + 1);
  W.resize(N + 1);
  for (ll i = 1; i <= N; ++i) cin >> V[i] >> W[i];
  I = min(N, (1ll << K) - 1);
  Dp.resize(I + 1);
  for (auto &dp : Dp) dp.resize(L + 1);
  for (ll i = 1; i <= I; ++i) {
    for (ll l = 0; l <= L; ++l) {
      Dp[i][l] = Dp[i / 2][l];
      if (W[i] <= l) Dp[i][l] = max(Dp[i][l], Dp[i / 2][l - W[i]] + V[i]);
    }
  }
  cin >> Q;
  for (ll q = 0; q < Q; ++q) {
    ll i, l;
    cin >> i >> l;
    cout << solve(i, l) << endl;
  }
  return 0;
}