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)

//header
#ifdef LOCAL
    #include "/Users/takakurashokichi/Desktop/atcoder/cxx-prettyprint-master/prettyprint.hpp"
    #define debug(x) cout << x << endl
#else
    #define debug(...) 42
#endif
    #pragma GCC optimize("Ofast")
    #include <bits/stdc++.h>
    //types
    using namespace std;
    using ll = long long;
    using ul = unsigned long long;
    using ld = long double;
    typedef pair < ll , ll > Pl;        
    typedef pair < int, int > Pi;
    typedef vector<ll> vl;
    typedef vector<int> vi;
    template< typename T >
    using mat = vector< vector< T > >;
    template< int mod >
    struct modint {
        int x;

        modint() : x(0) {}

        modint(int64_t y) : x(y >= 0 ? y % mod : (mod - (-y) % mod) % mod) {}

        modint &operator+=(const modint &p) {
            if((x += p.x) >= mod) x -= mod;
            return *this;
        }

        modint &operator-=(const modint &p) {
            if((x += mod - p.x) >= mod) x -= mod;
            return *this;
        }

        modint &operator*=(const modint &p) {
            x = (int) (1LL * x * p.x % mod);
            return *this;
        }

        modint &operator/=(const modint &p) {
            *this *= p.inverse();
            return *this;
        }

        modint operator-() const { return modint(-x); }

        modint operator+(const modint &p) const { return modint(*this) += p; }

        modint operator-(const modint &p) const { return modint(*this) -= p; }

        modint operator*(const modint &p) const { return modint(*this) *= p; }

        modint operator/(const modint &p) const { return modint(*this) /= p; }

        bool operator==(const modint &p) const { return x == p.x; }

        bool operator!=(const modint &p) const { return x != p.x; }

        modint inverse() const {
            int a = x, b = mod, u = 1, v = 0, t;
            while(b > 0) {
            t = a / b;
            swap(a -= t * b, b);
            swap(u -= t * v, v);
            }
            return modint(u);
        }

        modint pow(int64_t n) const {
            modint ret(1), mul(x);
            while(n > 0) {
            if(n & 1) ret *= mul;
            mul *= mul;
            n >>= 1;
            }
            return ret;
        }

        friend ostream &operator<<(ostream &os, const modint &p) {
            return os << p.x;
        }

        friend istream &operator>>(istream &is, modint &a) {
            int64_t t;
            is >> t;
            a = modint< mod >(t);
            return (is);
        }

        static int get_mod() { return mod; }
    };
    //abreviations
    #define all(x) (x).begin(), (x).end()
    #define rall(x) (x).rbegin(), (x).rend()
    #define rep_(i, a_, b_, a, b, ...) for (int i = (a), max_i = (b); i < max_i; i++)
    #define rep(i, ...) rep_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
    #define rrep_(i, a_, b_, a, b, ...) for (int i = (b-1), min_i = (a); i >= min_i; i--)
    #define rrep(i, ...) rrep_(i, __VA_ARGS__, __VA_ARGS__, 0, __VA_ARGS__)
    #define SZ(x) ((int)(x).size())
    #define pb(x) push_back(x)
    #define eb(x) emplace_back(x)
    #define mp make_pair
    #define print(x) cout << x << endl
    #define vsum(x) accumulate(x, 0LL)
    #define vmax(a) *max_element(all(a))
    #define vmin(a) *min_element(all(a))
    #define lb(c, x) distance((c).begin(), lower_bound(all(c), (x)))
    #define ub(c, x) distance((c).begin(), upper_bound(all(c), (x)))
    //functions
    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;}
    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 (b<a) { a=b; return 1; } return 0; }
    template< typename T >
    T mypow(T x, ll n) {
        T ret = 1;
        while(n > 0) {
            if(n & 1) (ret *= x);
            (x *= x);
            n >>= 1;
        }
        return ret;
    }
    ll modpow(ll x, ll n, const ll mod) {
        ll ret = 1;
        while(n > 0) {
            if(n & 1) (ret *= x);
            (x *= x);
            n >>= 1;
            x%=mod;
            ret%=mod;
        }
        return ret;
    }
    uint64_t my_rand(void) {
        static uint64_t x = 88172645463325252ULL;
        x = x ^ (x << 13); x = x ^ (x >> 7);
        return x = x ^ (x << 17);
    }
    //graph template
    template< typename T >
    struct edge {
        int src, to;
        T cost;

        edge(int to, T cost) : src(-1), to(to), cost(cost) {}

        edge(int src, int to, T cost) : src(src), to(to), cost(cost) {}

        edge &operator=(const int &x) {
            to = x;
            return *this;
        }
        operator int() const { return to; }
    };
    template< typename T >
    using Edges = vector< edge< T > >;
    template< typename T >
    using WeightedGraph = vector< Edges< T > >;
    using UnWeightedGraph = vector< vector< int > >;

//constant
//#define inf 1000000005LL
#define inf 4000000000000000005LL
#define mod 1000000007LL
#define endl '\n'
typedef modint<mod> mint;
const long double eps = 0.0001;
const long double PI  = 3.141592653589793;
//library
struct item {
  int w;
  int v;
  bool operator<(const item &r) const { return w < r.w; }
  void operator+=(const item &r) {
    w += r.w;
    v += r.v;
  }
};
using vt = vector<item>;
vt rekkyo(const vt& items){
    vt a, b, c;
    item x;
    x.w = 0, x.v = 0;
    a.pb(x);
    for(auto i:items){
        for(auto e:a){
            e+=i;
            b.pb(e);
        }
        merge(all(a), all(b), back_inserter(c));
        swap(a, c);
        b.clear();
        c.clear();
    }
    return a;
}
ll meet_in_the_middle(const vt& items, int l){
    int n = items.size();
    vt a = rekkyo({items.begin(), items.begin()+n/2});
    vt b = rekkyo({items.begin()+n/2, items.end()});
    reverse(all(a));
    int ans = 0;
    int maxv = -100000000;
    auto iter = b.begin();
    for(auto e:a){
        while (iter!=b.end()&&e.w+iter->w<=l)
        {
            chmax(maxv, iter->v);
            ++iter;
        }
        chmax(ans, e.v+maxv);
    }
    return ans;
}
int main(){
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << setprecision(20);
    int n; cin>>n;
    vt items(n);
    rep(i, n){
        cin>>items[i].v>>items[i].w;
    }
    int q; cin>>q;
    rep(_, q){
        int v, l; cin>>v>>l;
        vt tmp;
        while(v){
            tmp.pb(items[v-1]);
            v/=2;
        }
        print(meet_in_the_middle(tmp, l));
    }
}