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 <bitset>
#include <cassert>
#include <cctype>
#include <climits>
#include <cmath>
#include <complex>
#include <cstdio>
#include <cstring>
#include <deque>
#include <functional>
#include <iomanip>
#include <iostream>
#include <list>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <tuple>
#include <vector>
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
//#define cerr if(false) cerr
#ifdef DEBUG
#define show(...) cerr << #__VA_ARGS__ << " = ", debug(__VA_ARGS__);
#else
#define show(...) 42
#endif
using namespace std;
using ll = long long;
using pii = pair<int, int>;
template <typename T, typename S>
ostream& operator<<(ostream& os, pair<T, S> a) {
    os << '(' << a.first << ',' << a.second << ')';
    return os;
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> v) {
    for (auto x : v) os << x << ' ';
    return os;
}
void debug() {
    cerr << '\n';
}
template <typename H, typename... T>
void debug(H a, T... b) {
    cerr << a;
    if (sizeof...(b)) cerr << ", ";
    debug(b...);
}
int dp[1<<9][100001];
int v[1<<18], w[1<<18];
int solve(int V, int l){
    if(V < 1 << 9){
        return dp[V][l];
    }
    return max(solve(V / 2, l), solve(V / 2, l - w[V]) + v[V]);
}
int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;
    rep(i,n)cin >> v[i] >> w[i];
    rep(j,100001){
        if(j>=w[0])dp[0][j] = v[0];
    }
    rep(i,min(n, 1<<9) - 1){
        rep(j,100001){
            dp[i + 1][j] = max(dp[i/2][j], j - w[i + 1] >= 0 ? dp[i/2][j-w[i + 1]]+v[i + 1] : 0);
        }
    }
    int q;
    cin >> q;
    while(q--){
        int V, L;
        cin >> V >> L;
        cout << solve(V - 1, L) << "\n";
    }
}