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 vector<ll> vi;
typedef pair<ll,ll> pi;
typedef vector<pi> vpi;
typedef long double ld;
#define pb emplace_back
#define mp make_pair
#define lb lower_bound
#define ub upper_bound
#define ALL(x) x.begin(), x.end()
#define SZ(x) (ll)x.size()
#define f first
#define s second
const ll MAXN=266100;
const ll MAXK=19;
const ll INF = 1e15;
ll N,Q,a,b,c,d,e,f;
ll H = 9;

vpi V[MAXN];
pi T[MAXN];
vpi W[MAXN];
vpi queries[MAXN];
ll ans[MAXN];
ll J;
vpi tmp;

inline void fill(int i){
	vpi A;
	ll J = min(SZ(tmp), J);
	for (int bm=0;bm<(1<<J);++bm){
		pi cur = mp(0,0);
		for (int w=0;w<J;++w)if((1<<w)&bm){
			cur.f+=tmp[w].f;
			cur.s+=tmp[w].s;
		}
		A.pb(cur);
	}
	sort(ALL(A));
	for (auto t : A){
		if(SZ(V[i])){
			if(V[i].back().f == t.f){
				V[i].back().s = t.s;
			}else if(V[i].back().s >= t.s){
				continue;
			}
		}
		V[i].pb(t);
	}
}
int main(){
	ios_base::sync_with_stdio(0);cin.tie(0);
    cin>>N;
    ll layers = 0;
    ll r = N;
    while(r){
    	r/=2;
    	layers++;
    }
    H=min(H,layers);
   	for(int i=1;i<=N;++i){
   		cin>>a>>b;
   		T[i]=mp(b,a);
   	}
   	cin>>Q;
   	for(int i=0;i<Q;++i){
   		ans[i]=-1;
   		cin>>a>>b;
   		queries[a].pb(b,i);
   	}

   	V[0].pb(0,0);
   	c=1;
   	for (int i=1;i<=N;++i){
   		tmp.clear();
   		if (i<(1<<H)){
   			int cur=i;
	   		for (int j=0;j<H;++j){
	   			tmp.pb(T[cur]);
	   			cur=cur/2;
	   			if(cur==0)break;
	   		}
   			fill(i);
   			while(SZ(queries[i])){
   				ll l = queries[i].back().f;
   				int t = queries[i].back().s;
   				queries[i].pop_back();
   				pi x = *( --lb(ALL(V[i]), mp(l, INF)) );
   				ans[t]=x.s;
   			}
   		}else if (SZ(queries[i])){
   			int cur=i;
	   		for (int j=0;j<H;++j){
	   			tmp.pb(T[cur]);
	   			cur=cur/2;
	   			if(cur<(1<<H))break;
	   		}
   			fill(i);
   			int par = cur;
   			for (auto it : queries[i]){
   				ll l = it.f;
   				ll t = it.s;
   				// cerr<<"Query "<<i<<' '<<par<<'\n';
   				// cerr<<"Length "<<l<<'\n';
   				int ind = SZ(V[par])-1;
   				for (int x=0;x<SZ(V[i]);++x){
   					while(ind>-1&&V[par][ind].f+V[i][x].f > l){
   						--ind;
   					}
   					if(ind == -1)break;
   					ans[t] = max(ans[t], V[par][ind].s+V[i][x].s);
   				}
   				// exit(0);
   			}
   		}
   	}

   	for (int i=0;i<Q;++i){
   		cout<<ans[i]<<'\n';
   	}
}