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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;

inline int toInt(string s) {int v; istringstream sin(s);sin>>v;return v;}
template<class T> inline string toString(T x) {ostringstream sout;sout<<x;return sout.str();}

typedef long long ll;
typedef vector<int> VI;
typedef vector<VI> VVI;
typedef vector<long long> VL;
typedef vector<vector<long long>> VVL;
typedef vector<string> VS;
typedef pair<int, int> P;
typedef tuple<int,int,int> tpl;

#define ALL(a)  (a).begin(),(a).end()
#define SORT(c) sort((c).begin(),(c).end())
#define REVERSE(c) reverse((c).begin(),(c).end())
#define EXIST(m,v) (m).find((v)) != (m).end()
#define LB(a,x) lower_bound((a).begin(), (a).end(), x) - (a).begin()
#define UB(a,x) upper_bound((a).begin(), (a).end(), x) - (a).begin()

#define FOR(i,a,b) for(int i=(a);i<(b);++i)
#define REP(i,n)  FOR(i,0,n)
#define RFOR(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define RREP(i,n) RFOR(i,n,0)

#define en "\n"

constexpr double EPS = 1e-9;
constexpr double PI  = 3.1415926535897932;
constexpr int INF = 2147483647;
constexpr long long LINF = 1LL<<60;
constexpr long long MOD = 1000000007; // 998244353;

#define dump(x)  cerr << #x << " = " << (x) << endl;
#define debug(x) cerr << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

int N,Q;
VI V,W;
VVI item;
int dp[(1<<9)+1][100001];

void dfs(int v){
    if(v>=N) return;
    int par = (v-1)/2;
    if(v>0) item[v] = VI(item[par]);
    item[v].push_back(v);
    dfs(2*v+1); dfs(2*v+2);
    return;
}

void dfs2(int v){
    if(v>=N || item[v].size()>9) return;
    int par = v==0 ? 0 : (v-1)/2+1;

    REP(i,100000-W[v]+1){
        chmax(dp[v+1][i], dp[par][i]);
        chmax(dp[v+1][i+W[v]], dp[par][i]+V[v]);
    }

    dfs2(2*v+1); dfs2(2*v+2);
    return;
}

int solve(int v, int L){
    int K = item[v].size();
    if(K<=9) return dp[v+1][L];

    int par = (v-1)/2;
    while(par >= 1<<9){
        par = (par-1)/2;
    }
    par++;

    int N = K-9;
    int ans = 0;
    REP(i,1<<N){
        int vv = 0,ww = 0;
        REP(j,N) if(i & (1<<j)){vv += V[item[v][9+j]]; ww += W[item[v][9+j]];}
        if(ww>L) continue;
        vv += dp[par][L-ww];
        chmax(ans, vv);
    }
    return ans;
}

void Main(){
    cin >> N;
    V.resize(N); W.resize(N);
    REP(i,N) cin >> V[i] >> W[i];

    cin >> Q;
    item.resize(N);
    dfs(0); dfs2(0);
    REP(i,Q){
        int v,L;
        cin >> v >> L; v--;
        int ans = solve(v,L);
        cout << ans << en;
    }

    return;
}

int main(void){
    cin.tie(0);cout.tie(0);ios_base::sync_with_stdio(0);cout<<fixed<<setprecision(15);
    int t=1; //cin>>t;
    REP(_,t) Main();
    return 0;
}