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 <bits/stdc++.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rrep(i, n) for (int i = (int)(n - 1); i >= 0; i--)
#define FORR(i, a, b) for (int i = a; i >= b; --i)
#define all(x) (x).begin(), (x).end()
#define sz(x) int(x.size())
#define get_unique(x) x.erase(unique(all(x)), x.end());
typedef long long ll;
typedef complex<double> Complex;
const int INF = 1e9;
const ll MOD = 1e9 + 7;
const ll LINF = 1e18;
template <class T>
bool chmax(T& a, const T& b) {
    if (a < b) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
bool chmin(T& a, const T& b) {
    if (b < a) {
        a = b;
        return 1;
    }
    return 0;
}
template <class T>
vector<T> make_vec(size_t a) {
    return vector<T>(a);
}
template <class T, class... Ts>
auto make_vec(size_t a, Ts... ts) {
    return vector<decltype(make_vec<T>(ts...))>(a, make_vec<T>(ts...));
}
template <typename T>
ostream& operator<<(ostream& os, vector<T> v) {
    for (int i = 0; i < sz(v); i++) {
        os << v[i];
        if (i < sz(v) - 1) os << " ";
    }
    return os;
}
void calc(vector<ll> v, vector<ll> w, vector<pair<ll, ll>>& ret) {
    int n = v.size();
    ret.resize(1 << n);
    rep(i, 1 << n) {
        ll vv = 0, ww = 0;
        rep(j, n) {
            if ((i >> j) & 1) {
                vv += v[j];
                ww += w[j];
            }
        }
        ret[i] = (make_pair(ww, vv));
    }
    sort(all(ret));
}
int main() {
    int nn;
    cin >> nn;
    vector<ll> vv(nn + 1), ww(nn + 1);
    rep(i, nn) cin >> vv[i + 1] >> ww[i + 1];
    int q;
    cin >> q;
    vector<ll> v, w, v1, w1, v2, w2;
    vector<pair<ll, ll>> p1, p2;
    while (q--) {
        v.clear();
        w.clear();
        v1.clear();
        w1.clear();
        v2.clear();
        w2.clear();
        p1.clear();
        p2.clear();
        int now, l;
        cin >> now >> l;
        while (now) {
            v.push_back(vv[now]);
            w.push_back(ww[now]);
            now >>= 1;
        }
        int n = sz(v);
        rep(i, n) {
            if (i & 1) {
                v1.push_back(v[i]);
                w1.push_back(w[i]);
            } else {
                v2.push_back(v[i]);
                w2.push_back(w[i]);
            }
        }
        calc(v1, w1, p1);
        calc(v2, w2, p2);
        n = p1.size();
        int m = p2.size();
        ll ans = 0, ma = 0;
        int j = 0;
        FORR(i, n - 1, 0) {
            if (p1[i].first > l) continue;
            while (j < m && p1[i].first + p2[j].first <= l) {
                ma = max(ma, p2[j].second);
                j++;
            }
            ans = max(ans, p1[i].second + ma);
        }
        cout << ans << '\n';
    }
}