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>
#include <math.h>
#define rep(i,n) for (int i = 0; i < (n); ++i)
#define PI 3.14159265359
using namespace std;
using ll = long long;
using P = pair<int, int>;

ll dp[511][100001];
int w[1000000];
int v[1000000];

int main() {
  int n, q;
  cin >> n;
  rep(i, n) {
    cin >> v[i] >> w[i];
  }

  rep(i, 100001) {
    dp[0][i] = (i < w[0] ? 0 : v[0]);
  }

  for (int i=1; i<min(n, 511); i++) {
    rep(j, 100001) {
      dp[i][j] = max( dp[(i-1)/2][j], (j < w[i] ? 0 : dp[(i-1)/2][j-w[i]] + v[i]) );
    }
  }

  cin >> q;
  vector<P> query(q);
  rep(i, q) {
    cin >> query[i].first >> query[i].second;
  }
  rep(i, q) {
    int vid, l;
    vid = query[i].first - 1; l = query[i].second;

    if (vid < 511) {
      cout << dp[vid][l] << endl;
    }

    else {
      vector<int> wt;
      int tsize = 0, tpow = 1;
      int vidt = vid;
      while (vidt >= 511) {
        vidt = (vidt - 1) / 2;
        tsize++; tpow *= 2;
      }

      ll ans = 0;
      rep(j, tpow) {
        vidt = vid;
        ll vt = 0; int wt = 0;
        rep(k, tsize) {
          if ((j>>k) & 1) {
            vt += v[vidt]; wt += w[vidt];
          }
          vidt = (vidt - 1) / 2;
        }
        if (wt <= l) {
          ans = max( ans, vt + dp[vidt][l-wt] );
        }
      }
      cout << ans << endl;
    }
  }

/*  rep(i, min(n, 511)) {
    rep(j, 7) cout << dp[i][j] << " ";
    cout << endl;
  }*/

  return 0;
}

/*
*/