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: Runtime Error)

#include <bits/stdc++.h>
#define rep(i,n) for(int i=0;i<n;++i)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
#define reps(i,a,b) for(int i=a;i<b;++i)
#define fi first
#define se second
#define pb push_back
#define eb emplace_back
#define sz(x) (int)(x).size()
#define all(v) v.begin(),v.end()
#define rall(v) v.rbegin(),v.rend()
using namespace std;
using ll = long long;
using P = pair<int, int>;
template<typename T> inline bool chmax(T &a,T b){if(a<b){a=b;return 1;}return 0;}
template<typename T> inline bool chmin(T &a,T b){if(b<a){a=b;return 1;}return 0;}
template<typename T> vector<T> make_vec(size_t a){return vector<T>(a);}
template<typename T,typename... Ts>
auto make_vec(size_t a,Ts... ts){return vector<decltype(make_vec<T>(ts...))>(a,make_vec<T>(ts...));}
template<typename T,typename U,typename... V>
typename enable_if<is_same<T,U>::value>::type fill_v(U &u,const V... v){u=U(v...);}
template<typename T,typename U,typename... V>
typename enable_if<!is_same<T,U>::value>::type fill_v(U &u,const V... v){for(auto &e:u)fill_v<T>(e,v...);}

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int n;
    cin >> n;
    vector<int> v(n+1), w(n+1);
    rep(i, n) cin >> v[i+1] >> w[i+1];
    int q;
    cin >> q;
    auto dp = make_vec<int>(n+1, 100001);
    vector<bool> seen(n+1);
    seen[0] = true;
    while (q--) {
        int u, W;
        cin >> u >> W;
        int s = u;
        vector<int> child;
        while (!seen[s]) {
            child.pb(s);
            s /= 2;
        }
        for (int i = sz(child)-1; i >= 0; --i) {
            int k = child[i];
            rep(j, 100001) {
                chmax(dp[k][j], dp[k/2][j]);
                if (j != 0) chmax(dp[k][j], dp[k][j-1]);
                if (j+w[k] <= 100000) chmax(dp[k][j+w[k]], dp[k/2][j] + v[k]);
            }
            seen[k] = true;
        }
        cout << dp[u][W] << endl;
    }
}