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 x first
#define y second
#define mp make_pair
#define pb push_back

typedef long long ll;

const ll MOD = 1e9+7;
const ll INF = 1e9+5;

int v[300005];
int w[300005];

ll memo[1055][100005];

ll dp(int pos, int l) {
	if (pos == 0) return 0;
	if (memo[pos][l] != -1) return memo[pos][l];
	if (l < w[pos]) return memo[pos][l] = dp(pos/2, l);
	return memo[pos][l] = max(dp(pos/2, l), dp(pos/2, l-w[pos])+v[pos]);
}

ll rek(int pos, int l) {
	if (pos < 1024) return dp(pos, l);
	if (l < w[pos]) return rek(pos/2, l);
	return max(rek(pos/2, l), rek(pos/2, l-w[pos])+v[pos]);
}

int main() {
	memset(memo, -1, sizeof memo);
	int n;
	scanf("%d", &n);
	for (int i=1 ; i<=n ; i++) {
		scanf("%d%d", v+i, w+i);
	}
	int q;
	scanf("%d", &q);
	for (int i=0 ; i<q ; i++) {
		int pos, l;
		scanf("%d%d", &pos, &l);
		printf("%lld\n", rek(pos, l));
	}
	return 0;
}