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>
using namespace std;
using ll=long long;
const ll MOD=(ll)1e9+7;
//const ll MOD=(ll)998244353;
const ll INF=(ll)1e18+10;
const double PI = 2.0*acos(0);
template<class T> bool chmax(T &a, const T &b) { if (a<b) { a=b; return 1; } return 0; }
template<class T> bool chmin(T &a, const T &b) { if (a>b) { a=b; return 1; } return 0; }
int dy[]={-1,1,0,0};
int dx[]={0,0,-1,1};



void solve(int ca){
    int N; cin>>N;
    vector<int> v(N+1),w(N+1);
    for (int i = 1; i <= N; ++i) {
        cin>>v[i]>>w[i];
    }

    int Q; cin>>Q;
    vector<int> u(N),L(N);
    int maxiL=0;
    for (int i = 0; i < Q; ++i) {
         cin>>u[i]>>L[i];
         chmax(maxiL,L[i]);
    }
    //maxiL=500010;
    int half = (1<<((int)ceil(log2(N))/2))-1;
    //cout<<half<<endl;
    vector<vector<ll>> dp(half+1,vector<ll>(maxiL+1,0));

    for (int i = 1; i <= half; ++i) {
        int par=i/2;
        for (int j = 0; j <= maxiL; ++j) {
            dp[i][j]=dp[par][j];
            if(j-w[i]>=0) chmax(dp[i][j],dp[par][j-w[i]]+v[i]);
        }
    }
    /*
    for(auto i:dp){
        for(auto j:i){
            cout<<j<<" ";
        }
        cout<<endl;
    }
     */


    for (int i = 0; i < Q; ++i) {
        if(u[i]<=half){
            printf("%lld\n",dp[u[i]][L[i]]);
        }else{
            vector<int> nodes;
            int n=u[i];
            while(n>half){
                nodes.push_back(n);
                n>>=1;
            }
            const int M=nodes.size();
            ll ans=0;
            for (int j = 0; j < (1 << M); ++j) {
                ll vsum=0, wsum=0;
                for (int k = 0; k < M; ++k) {
                    if((j>>k)&1){
                        vsum+=v[nodes[k]];
                        wsum+=w[nodes[k]];
                    }
                }
                //cout<<"called"<<endl;
                if(wsum>L[i]) continue;
                chmax(ans,dp[n][L[i]-wsum]+vsum);
            }
            cout<<ans<<endl;
        }

    }

}

//#define MULTI
int main() {
#ifdef MULTI
    int N; cin>>N;
    for (int i = 0; i < N; ++i) {
        solve(i+1);
    }
#else
    solve(0);
#endif
    return 0;
}