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;
typedef long long ll;
typedef pair<ll, ll> p_ll;

template<class T>
void debug(T itr1, T itr2) { auto now = itr1; while(now<itr2) { cout << *now << " "; now++; } cout << endl; }
#define repr(i,from,to) for (int i=(int)from; i<(int)to; i++)
#define all(vec) vec.begin(), vec.end()
#define rep(i,N) repr(i,0,N)
#define per(i,N) for (int i=(int)N-1; i>=0; i--)

const ll MOD = pow(10,9)+7;
const ll LLINF = pow(2,61)-1;
const int INF = pow(2,30)-1;

vector<ll> fac;
void c_fac(int x=pow(10,6)+10) { fac.resize(x,true); rep(i,x) fac[i] = i ? (fac[i-1]*i)%MOD : 1; }
ll inv(ll a, ll m=MOD) { ll b = m, x = 1, y = 0; while (b!=0) { int d = a/b; a -= b*d; swap(a,b); x -= y*d; swap(x,y); } return (x+m)%m; }
ll nck(ll n, ll k) { return fac[n]*inv(fac[k]*fac[n-k]%MOD)%MOD; }
ll gcd(ll a, ll b) { if (a<b) swap(a,b); return b==0 ? a : gcd(b, a%b); }
ll lcm(ll a, ll b) { return a/gcd(a,b)*b; }

struct vertex { ll v, w; };
struct query { ll v, L; };
vector<vertex> v;

ll MA = 100000, D = 9;

int main() {
  ll N; cin >> N;
  v.resize(1<<18); rep(i,N) cin >> v[i+1].v >> v[i+1].w;
  ll Q; cin >> Q;
  query q[Q]; rep(i,Q) cin >> q[i].v >> q[i].L;

  vector<vector<ll>> c(1<<D,vector<ll>(MA,0));
  rep(i,1<<D) repr(j,1,MA) {
    if (j<v[i].w) c[i][j] = max(c[i][j-1], c[i/2][j]);
    else c[i][j] = max(c[i/2][j], c[i/2][j-v[i].w]+v[i].v);
  }
  // rep(i,10) debug(c[i].begin(),c[i].begin()+10);


  rep(qu,Q) {
    ll dig = 0, p2 = 1; while (p2<=q[qu].v) { dig++; p2*=2; }
    ll result = 0;
    if (dig<=D) result = c[q[qu].v][q[qu].L];
    else {
      // cout << q[qu].v << " " << q[qu].L << endl;
      vertex num[1<<dig-D] = {};
      rep(i,dig-D) rep(j,1<<i) {
        num[(1<<i)+j].v = num[j].v+v[q[qu].v>>i].v;
        num[(1<<i)+j].w = num[j].w+v[q[qu].v>>i].w;
      }
      // rep(i,1<<dig-D) cout << num[i].w << "->" << num[i].v << " "; cout << endl;
      rep(i,1<<dig-D) if (num[i].w<=q[qu].L) result = max(result, c[q[qu].v>>dig-D][q[qu].L-num[i].w]+num[i].v);
    }
    cout << result << endl;
  }
  return 0;
}