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<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/tree_policy.hpp>
using namespace __gnu_pbds;
using namespace std;
#define int long long int
// #define ll int128_t
#define ld long double
#define ff first
#define ss second
#define mp make_pair
#define IOS std::ios::sync_with_stdio(false); cin.tie(NULL); cout.tie(NULL);
#define pb push_back
#define eb emplace_back
#define pii pair<int, int>
#define endl '\n'
#define all(x) x.begin(),x.end()
#define f(i,j,l) for(i=j;i<l;i++)
#define smax(a,b) a=max(a,b)
// #if !defined(ONLINE_JUDGE)
// #include "debugging.h"
// #endif
template <typename T>
using ordered_set = tree<T, null_type, less<T>, rb_tree_tag, tree_order_statistics_node_update>;
// find_by_order(k)  returns iterator to kth element starting from 0;
// order_of_key(k) returns count of elements strictly smaller than k;
// erase,insert same as normal set

// int const mod = 7340033;
int const mod = 998244353;
int power(int a,int b,int m){
    int res = 1;
    while(b){
        if(b&1){
            res = res*a;
            res %= m;
        }
        a=a*a;
        a%=m;
        b=b/2;
    }
    return res;
}

int inverse(int x){
    return power(x,mod-2,mod);
}

int const N = 3e5+55;
int const inf = 2e16;

int val[N] , w[N];
int dp[(1LL<<9LL)][N];

int height(int n){
    for(int i=17;i>=0;i--){
        if(n&(1LL<<i)) return i+1; 
    }
}

void solve(){
    int i,j,l;  
    int n;
    cin >> n;
    for(i=1;i<=n;i++){
        cin >> val[i] >> w[i];
    }

    int k = height(n)/2;

    for(i=1;i<(1LL<<k);i++){
        int h = height(i);
        for(j=0;j<(1LL<<h);j++){
            int vv = 0;
            int ww = 0;
            for(l=0;l<h;l++){
                if(j&(1LL<<l)){
                    vv += val[i>>l];
                    ww += w[i>>l];
                }
            }
            if(ww < N){
                dp[i][ww] = max(dp[i][ww] , vv);
            }
        }

        for(j=1;j<N;j++){
            dp[i][j] = max(dp[i][j] , dp[i][j-1]);
            // cout<<dp[i][j]<<" ";
        }
    }

    int q;
    cin>>q;
    while(q--){
        int node,lim;
        cin >> node >> lim;
        int h = height(node);
        if(h <= k){
            cout<<dp[node][lim]<<endl;
        }
        else{
            int rem = h-k;
            int ans = 0;
            for(j=0;j<(1LL<<rem);j++){
                int vv = 0;
                int ww = 0;

                for(l=0;l<rem;l++){
                    if(j&(1LL<<l)){
                        vv += val[node>>l];
                        ww += w[node>>l];
                    }
                }

                // cout<<node<<" "<<vv<<" "<<ww<<" "<<(node>>rem) << endl;

                if(ww <= lim){
                    ans = max(ans , vv + dp[node>>rem][lim-ww]);
                }
            }
            cout<<ans<<endl;
        }
    }
}               


int32_t main(){
    IOS;
    int t;
    t = 1;
    // precompute();
    // cin>>t;
    for(int tt=1;tt<=t;tt++){
        solve();
    }
    return 0;
}


/*


*/