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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;
#define all(x)   (x).begin(),(x).end()
#define YES() printf("YES\n")
#define NO() printf("NO\n")
#define isYES(x) printf("%s\n",(x) ? "YES" : "NO")
#define Yes() printf("Yes\n")
#define No() printf("No\n")
#define isYes(x) printf("%s\n",(x) ? "Yes" : "No")
#define isIn(x,y,h,w) (x >= 0 && x < h && y >= 0 && y < w)

#define int long long
//using ll = long long;
using P = pair<int,int>;

template<class T> T &chmin(T &a,const T &b){ return a = min(a,b); }
template<class T> T &chmax(T &a,const T &b){ return a = max(a,b); }
 
const int INF=1e+18;
const double EPS=1e-9;
const int MOD=1000000007;

const int dx[]={1,0,-1,0},dy[]={0,-1,0,1};

ostream &operator<<(ostream &os,const P &p){ return os << "(" << p.first << "," << p.second << ")"; }
template<class T>
ostream &operator<<(ostream &os,const vector<T> &v){
	for(int i = 0;i < v.size();i++) os << (i ? "," : "[") << v[i];
	os << "]";
	return os;
}

int depth(int v){
	int ret = 0;
	while(v){
		ret++;
		v = (v - 1) / 2;
	}
	return ret;
}

const int L_LIM = 100010,D_MAX = 9;

vector<P> rekkyo(vector<int> ws,vector<int> vs){
	vector<P> ans;
	int d = ws.size();
	for(int i = 0;i < (1 << d);i++){
		int sumv = 0,sumw = 0;
		for(int j = 0;j < d;j++){
			if(i >> j & 1){
				sumv += vs[j];
				sumw += ws[j];
			}
		}
		if(sumw <= L_LIM) ans.emplace_back(sumw,sumv);
	}
	sort(all(ans));
	int ma = 0;
	for(int i = 0;i < ans.size();i++){
		chmax(ma,ans[i].second);
		ans[i].second = ma;
	}
	return ans;
}

signed main(){
	int n,q,v[1 << 18],w[1 << 18];
	vector<P> ans[1 << 9];
	cin >> n;
	for(int i = 0;i < n;i++){
		cin >> v[i] >> w[i];
	}
	for(int i = 0;i < n;i++){
		int pos = i,d = depth(i),cnt = 0;
		if(depth(i) >= D_MAX) break;
		vector<int> ws(d + 1),vs(d + 1);
		ws[0] = w[pos];
		vs[0] = v[pos];
		while(pos){
			pos = (pos - 1) / 2;
			ws[++cnt] = w[pos];
			vs[cnt] = v[pos];
		}
		//cout << ws << vs << endl;
		ans[i] = rekkyo(ws,vs);
		//cout << ans[i] << endl;
	}
	cin >> q;
	for(int i = 0;i < q;i++){
		int x,l;
		cin >> x >> l; x--;
		int d = depth(x),pos = x,ma = 0;
		vector<int> ws(max(0ll,d - D_MAX + 1)),vs(max(0ll,d - D_MAX + 1));
		for(int j = 0;j < d - (D_MAX - 1);j++){
			ws[j] = w[pos];
			vs[j] = v[pos];
			pos = (pos - 1) / 2;
		}
		//cout << ws << vs << endl;
		auto rek = rekkyo(ws,vs);
		for(P p : rek){
			int a = upper_bound(all(ans[pos]),P(l - p.first,INF)) - ans[pos].begin() - 1;
			if(a >= 0) chmax(ma,ans[pos][a].second + p.second);
		}
		cout << ma << endl;
	}
}