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: Accepted)

#include <bits/stdc++.h>
#include <cstdlib>
#include <cmath>
#include <algorithm>
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using Graph= vector<vector<ll>>;
struct edge{ll to ; ll cost ;} ;
using graph =vector<vector<edge>> ;
#define rep(i,n) for (ll i=0; i < (n); ++i)
#define rep2(i,n,m) for(ll i=n;i<=m;i++)
#define rep3(i,n,m) for(ll i=n;i>=m;i--)
#define pb push_back
#define eb emplace_back
#define ppb pop_back
#define mpa make_pair
#define fi first
#define se second 
#define set20 cout<<fixed<<setprecision(20) ;
const ll INF=1e18 ;
inline void chmax(ll& a,ll b){a=max(a,b);}
inline void chmin(ll& a,ll b){a=min(a,b);}
double pi=acos(-1) ;
ll gcd(ll a, ll b) { return b?gcd(b,a%b):a;}
ll lcm(ll a, ll b) { return a/gcd(a,b)*b;}
 
ll dp[1025][100005] ;
 int main(){
 ll n ; cin>> n ;
 vector<ll> v(n) ,w(n) ;
 rep(i,n) cin>>v[i]>>w[i] ;
 ll q ; cin>> q ;
rep(i,100005){
  dp[0][i]=0;
}
rep2(i,1,min(1023ll,n)){
 rep(j,100005){
   if(j>=w[i-1]) dp[i][j]=max(dp[i/2][j],dp[i/2][j-w[i-1]]+v[i-1]) ;
   else dp[i][j]= dp[i/2][j] ;
 }
}

 while(q--){
  ll a, b ; cin>>a>>b ;
  if(a<=1023) cout<<dp[a][b]<<endl ;
  
  else{
    vector<ll> koho ;
    while(a>1023){
      koho.pb(a) ;
      a/=2 ;
    }
    ll ans= dp[a][b] ;
    ll len = koho.size() ;
   rep(i,1<<len){
     bitset<9> s(i) ;
     ll we=0 ;
     ll va=0 ;
     rep(j,9){
       if(s[j]) {
         we+= w[koho[j]-1] ;
         va+= v[koho[j]-1] ;
       }
       }
     if(we<=b) chmax(ans,dp[a][b-we]+va) ;

   } 
      cout<<ans<<endl ;
  }
}


   return 0 ;
 }