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 <bits/stdc++.h>
using namespace std;

#define int long long
#define rep(i, n) for (int i = 0; i < (int) (n); i++)
#define reps(i, n) for (int i = 1; i <= (int) (n); i++)
#define all(x) (x).begin(), (x).end()
#define uniq(x) (x).erase(unique(all(x)), (x).end()))
#define bit(n) (1LL << (n))
#define cdiv(a, b) (((a) - 1) / (b) + 1)
#define dump(x) cerr << #x " = " << (x) << endl
using vint = vector<int>;
using vvint = vector<vint>;
using pint = pair<int, int>;
using vpint = vector<pint>;
template<typename T> using priority_queue_rev = priority_queue<T, vector<T>, greater<T>>;
constexpr double PI = 3.1415926535897932384626433832795028;
constexpr int DY[8] = {0, 1, 0, -1, 1, 1, -1, -1};
constexpr int DX[8] = {1, 0, -1, 0, 1, -1, -1, 1};
int gcd(int a, int b) {
    while (b) { swap(a %= b, b); }
    return a;
}
int lcm(int a, int b) { return a / gcd(a, b) * b; }
template<typename T> void fin(T mes) {
    cout << mes << endl;
    exit(0);
}
template<typename T, typename U> bool chmax(T &a, const U &b) {
    if (a < b) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> bool chmin(T &a, const U &b) {
    if (b < a) {
        a = b;
        return true;
    }
    return false;
}
template<typename T, typename U> ostream &operator<<(ostream &os, const pair<T, U> &rhs) {
    os << "(" << rhs.first << ", " << rhs.second << ")";
    return os;
}
template<typename T> ostream &operator<<(ostream &os, const vector<T> &rhs) {
    os << "{";
    for (auto itr = rhs.begin(); itr != rhs.end(); itr++) {
        os << *itr << (next(itr) != rhs.end() ? ", " : "");
    }
    os << "}";
    return os;
}
struct setup {
    static constexpr int PREC = 20;
    setup() {
        cout << fixed << setprecision(PREC);
        cerr << fixed << setprecision(PREC);
    };
} setup;

signed main() {
    int N;
    cin >> N;
    vint V(N + 1), W(N + 1);
    reps(i, N) { cin >> V[i] >> W[i]; }
    int Q;
    cin >> Q;
    rep(i, Q) {
        int v, L;
        cin >> v >> L;
        vint Vc, Wc;
        while (v > 0) { Vc.push_back(V[v]), Wc.push_back(W[v]), v >>= 1; }
        map<int, int> WV;
        int n = Vc.size() / 2;
        rep(j, bit(n)) {
            int Vcc = 0, Wcc = 0;
            rep(k, n) { if (j & bit(k)) { Vcc += Vc[k], Wcc += Wc[k]; }}
            chmax(WV[Wcc], Vcc);
        }
        int c = 0;
        for (auto p:WV) {
            chmax(c, p.second);
            p.second = c;
        }
        int ans = 0;
        rep(j, bit(Vc.size() - n)) {
            int Vcc = 0, Wcc = 0;
            rep(k, Vc.size() - n) { if (j & bit(k)) { Vcc += Vc[k + n], Wcc += Wc[k + n]; }}
            if (Wcc > L) { continue; }
            chmax(ans, prev(WV.upper_bound(L - Wcc))->second + Vcc);
        }
        cout << ans << endl;
    }
}