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;
using lint = long long int;
using P = pair<int, int>;
#define FOR(i, begin, end) for(int i=(begin),i##_end_=(end);i<i##_end_;i++)
#define IFOR(i, begin, end) for(int i=(end)-1,i##_begin_=(begin);i>=i##_begin_;i--)
#define REP(i, n) FOR(i,0,n)
#define IREP(i, n) IFOR(i,0,n)
#define ALL(a)  (a).begin(),(a).end()
constexpr int MOD = 1000000007;
constexpr int INF = 2147483647;
void yes(bool expr) {
  cout << (expr ? "Yes" : "No") << "\n";
}
struct Node;
vector<Node> F;
int N;
int maxd = 9;
lint calc = 0;
struct Node {
  int idx;
  int v;
  int w;
  int d = 0;
  vector<int> dp;
  void calcd() {
      int tmp = 2;
      while(idx >= tmp) {
          tmp *= 2;
          d++;
      }
  }
  void precalc(vector<int> dpt) {
      if(d > maxd) return;
      IREP(i, 100001) if(i-w >= 0) dpt[i] = max(dpt[i], dpt[i-w]+v);
      FOR(i, 1, 100001) dpt[i] = max(dpt[i], dpt[i-1]);
      calc += 200000;
      dp = dpt;
      if(idx*2-1 < N) F[idx*2-1].precalc(dp);
      if(idx*2 < N) F[idx*2].precalc(dp);
  }
  int dfs(vector<int> &V, vector<int> &W) {
      //cout << idx << endl;
      if(d <= maxd) return idx;
      calc++;
      V.push_back(v);
      W.push_back(w);
      return F[idx/2-1].dfs(V, W);
  }
};
int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
    cin >> N;
    REP(i, N) {
        F.emplace_back();
        F[i].idx = i+1;
        cin >> F[i].v >> F[i].w;
        //F[i].v = 1; F[i].w = 1;
        F[i].calcd();
    }
    vector<int> dp(100001);
    F[0].precalc(dp);

    //cout << calc << endl;

    int Q;
    cin >> Q;
    REP(q, Q) {
        int v, L;
        cin >> v >> L;
        //v = 249999; L = 100000;

        if(F[v-1].d <= maxd) {
            cout << F[v-1].dp[L] << "\n";;
            continue;
        }

        vector<int> V = {0};
        vector<int> W = {0};
        vector<int> dp = F[F[v-1].dfs(V, W)-1].dp;

        //REP(i, V.size()) cout << V[i] << (i!=V.size()-1 ? " " : "");
        //cout << "\n";

        int ans = 0;
        int Vt[1100];
        int Wt[1100];
        Vt[0] = 0;
        Wt[0] = 0;
        int vtsize = 1;
        int idx = 1;
        REP(i, V.size()) {
            REP(j, vtsize) {
                int vsum = V[i] + Vt[j];
                int wsum = W[i] + Wt[j];
                if(wsum <= L && dp[L-wsum]+vsum > ans) ans = dp[L-wsum] + vsum;
                Vt[idx] = vsum;
                Wt[idx] = wsum;
                idx++;
                calc++;
            }
            vtsize *= 2;
        }
        //cout << idx << endl;

        cout << ans << "\n";
        
    }

    //cout << calc << endl;
    

}