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 Go (1.14.1) (judge verdict: Runtime Error)

package main

import (
	"bufio"
	"fmt"
	"os"
	"strconv"
)

var sc = bufio.NewScanner(os.Stdin)

func Scanner() string {
	sc.Scan()
	return sc.Text()
}

func intScanner() int {
	n, _ := strconv.Atoi(Scanner())
	return n
}

func floatScanner() float64 {
	n, _ := strconv.ParseFloat(Scanner(), 64)
	return n
}
func stringToInt(s string) int {
	n, _ := strconv.Atoi(s)
	return n
}
func stringSliceScanner(n int) []string {
	a := make([]string, n)
	for i := 0; i < n; i++ {
		a[i] = Scanner()
	}
	return a
}
func intSliceScanner(n int) []int {
	a := make([]int, n)
	for i := 0; i < n; i++ {
		a[i] = intScanner()
	}
	return a
}
func larger(a, b int) int {
	if a < b {
		return b
	} else {
		return a
	}
}
func smaller(a, b int) int {
	if a > b {
		return b
	} else {
		return a
	}
}
func max(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x < a[i] {
			x = a[i]
		}
	}
	return x
}
func min(a []int) int {
	x := a[0]
	for i := 0; i < len(a); i++ {
		if x > a[i] {
			x = a[i]
		}
	}
	return x
}
func abs(x int) int {
	if x < 0 {
		x = -x
	}
	return x
}

var mod int

func main() {
	mod = 1000000007
	buf := make([]byte, 0)
	sc.Buffer(buf, mod)
	sc.Split(bufio.ScanWords)
	n := intScanner()
	v := make([]int, n+1)
	w := make([]int, n+1)
	for i := 1; i <= n; i++ {
		v[i] = intScanner()
		w[i] = intScanner()
	}
	dp := make([][]int, n+1)
	dp[0] = make([]int, 100001)
	for i := 1; i <= n; i++ {
		dp[i] = make([]int, 100001)
		for j := 0; j < 100001; j++ {
			if j >= w[i] {
				dp[i][j] = larger(dp[i/2][j], dp[i/2][j-w[i]]+v[i])
			} else {
				dp[i][j] = dp[i/2][j]
			}
		}
	}
	q := intScanner()
	for i := 0; i < q; i++ {
		x, l := intScanner(), intScanner()
		fmt.Println(dp[x][l])
	}
}