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: Time Limit Exceeded)

package main

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

func configure(scanner *bufio.Scanner) {
	scanner.Split(bufio.ScanWords)
	scanner.Buffer(make([]byte, 1000005), 1000005)
}
func getNextString(scanner *bufio.Scanner) string {
	scanned := scanner.Scan()
	if !scanned {
		panic("scan failed")
	}
	return scanner.Text()
}
func getNextInt(scanner *bufio.Scanner) int {
	i, _ := strconv.Atoi(getNextString(scanner))
	return i
}
func getNextInt64(scanner *bufio.Scanner) int64 {
	i, _ := strconv.ParseInt(getNextString(scanner), 10, 64)
	return i
}
func getNextFloat64(scanner *bufio.Scanner) float64 {
	i, _ := strconv.ParseFloat(getNextString(scanner), 64)
	return i
}
func main() {
	fp := os.Stdin
	wfp := os.Stdout
	extra := 0
	if os.Getenv("I") == "IronMan" {
		fp, _ = os.Open(os.Getenv("END_GAME"))
		extra = 100
	}
	scanner := bufio.NewScanner(fp)
	configure(scanner)
	writer := bufio.NewWriter(wfp)
	defer func() {
		r := recover()
		if r != nil {
			fmt.Fprintln(writer, r)
		}
		writer.Flush()
	}()
	solve(scanner, writer)
	for i := 0; i < extra; i++ {
		fmt.Fprintln(writer, "-----------------------------------")
		solve(scanner, writer)
	}
}
func solve(scanner *bufio.Scanner, writer *bufio.Writer) {
	n := getNextInt(scanner)
	vv := make([]int64, n+1)
	ww := make([]int64, n+1)
	for i := 0; i < n; i++ {
		vv[i+1] = getNextInt64(scanner)
		ww[i+1] = getNextInt64(scanner)
	}
	q := getNextInt(scanner)
	for i := 0; i < q; i++ {
		v := getNextInt(scanner)
		l := getNextInt64(scanner)
		ans := int64(0)
		bb := make([]int, 0)
		for b := v; b > 0; b /= 2 {
			bb = append(bb, b)
		}
		bl := len(bb)
		pp1 := make([]pair, 0)
		pp1 = append(pp1, pair{0, 0})
		for k := 0; k < bl/2; k++ {
			pp1 = update(ww[bb[k]], vv[bb[k]], pp1)
		}
		pp2 := make([]pair, 0)
		pp2 = append(pp2, pair{0, 0})
		for k := 0; k < bl-bl/2; k++ {
			pp2 = update(ww[bb[bl/2+k]], vv[bb[bl/2+k]], pp2)
		}

		rr := len(pp2) - 1
		for j := 0; j < len(pp1); j++ {
			for rr >= 0 && pp1[j][0]+pp2[rr][0] > l {
				rr--
			}
			if rr < 0 {
				break
			}
			if ans < pp1[j][1]+pp2[rr][1] {
				ans = pp1[j][1] + pp2[rr][1]
			}
		}
		fmt.Fprintln(writer, ans)
	}

}

type pair [2]int64

func update(w, v int64, pp []pair) []pair {
	pp2 := make([]pair, len(pp))
	pp3 := make([]pair, len(pp)<<1)
	for i := 0; i < len(pp); i++ {
		pp2[i][0] = pp[i][0] + w
		pp2[i][1] = pp[i][1] + v
	}
	var i, j, k int
	for i < len(pp) || j < len(pp) {
		if i == len(pp) {
			pp3[k][0] = pp2[j][0]
			pp3[k][1] = pp2[j][1]
			j++
			k++
			continue
		}
		if j == len(pp) {
			pp3[k][0] = pp[i][0]
			pp3[k][1] = pp[i][1]
			i++
			k++
			continue
		}
		if pp[i][0] == pp2[j][0] {
			if pp[i][1] > pp2[j][1] {
				pp3[k][0] = pp2[j][0]
				pp3[k][1] = pp2[j][1]
				j++
				k++
				continue
			}
			pp3[k][0] = pp[i][0]
			pp3[k][1] = pp[i][1]
			i++
			k++
			continue
		}
		if pp[i][0] < pp2[j][0] {
			pp3[k][0] = pp[i][0]
			pp3[k][1] = pp[i][1]
			i++
			k++
			continue
		}
		pp3[k][0] = pp2[j][0]
		pp3[k][1] = pp2[j][1]
		j++
		k++
	}
	for i := 1; i < len(pp3); i++ {
		if pp3[i][1] < pp3[i-1][1] {
			pp3[i][1] = pp3[i-1][1]
		}
	}
	return pp3
}