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: Wrong Answer)

#include <algorithm>
#include <iostream>
#include <vector>

using namespace std;
typedef long long ll;
ll dp[520][100001];
ll n, q, v[100001], w[100001];

ll dfs(ll x, ll y, ll vv, ll ww) {
  if (x < (1 << 8)) {
    if (ww > y) {
      return 0LL;
    }
    return dp[x][y - ww] + vv;
  }
  return max(dfs(x / 2, y, vv + v[x], ww + w[x]), dfs(x / 2, y, vv, ww));
}

int main() {
  ll vv, l;
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> v[i + 1] >> w[i + 1];
  }
  for (ll i = 1LL; i <= min(n, (1LL << 8)); i++) {
    for (ll j = 100000LL; j >= 0; j--) {
      dp[i][j] = dp[i / 2LL][j];
      if (j - w[i] >= 0) {
        dp[i][j] = max(dp[i][j], dp[i / 2][j - w[i]] + v[i]);
      }
    }
    for (ll j = 1; j <= 100000; j++) {
      dp[i][j] = max(dp[i][j], dp[i][j - 1]);
    }
  }
  cin >> q;
  for (int i = 0; i < q; i++) {
    cin >> vv >> l;
    cout << dfs(vv, l, 0, 0) << endl;
  }
}