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"
  "os"
  "strconv"
  "strings"
  "fmt"
)

func Solve() {
  var N, Q, v, L int
  NextInt(&N)
  V := make([]int, N + 1)
  W := make([]int, N + 1)
  for i := 1; i <= N; i++ { NextInt(&V[i], &W[i]) }
  U := MinInt(N + 1, 1 << 9)
  DP := make([][]int, U)
  L = 100001
  for i := range DP { DP[i] = make([]int, L) }
  for v := 1; v < U; v++ {
    for j, dp := range DP[v / 2] {
      if j >= W[v] {
        DP[v][j] = MaxInt(DP[v / 2][j - W[v]] + V[v], dp)
      } else {
        DP[v][j] = dp
      }
    }
  }
  var F func(int, int) int
  F = func(v, L int) int {
    if v < U { return DP[v][L] }
    a := F(v / 2, L)
    if W[v] <= L { a = MaxInt(F(v / 2, L - W[v]) + V[v], a) }
    return a
  }
  NextInt(&Q)
  for i := 0; i < Q; i++ {
    NextInt(&v, &L)
    Write(F(v, L))
  }
}

func main() {
  Solve()
  Output()
}

var reader = bufio.NewReaderSize(os.Stdin, 1000000)
var writer = bufio.NewWriter(os.Stdout)
func NextLine() string {
  buffer := make([]byte, 0)
  for true {
    line, isPrefix, err := reader.ReadLine()
    if err != nil { panic(err) }
    buffer = append(buffer, line...)
    if !isPrefix { break }
  }
  return string(buffer)
}
func NextInt(A ...*int) {
  L := strings.Split(NextLine(), " ")
  for i, a := range A {
    *a, _ = strconv.Atoi(L[i])
  }
}
func NextIntVec(A *[]int) {
  L := strings.Split(NextLine(), " ")
  (*A) = make([]int, len(L))
  for i, l := range L {
    (*A)[i], _ = strconv.Atoi(l)
  }
}
func NextFloat(A ...*float64) {
  L := strings.Split(NextLine(), " ")
  for i, a := range A {
    *a, _ = strconv.ParseFloat(L[i], 64)
  }
}
func NextFloatVec(A *[]float64) {
  L := strings.Split(NextLine(), " ")
  (*A) = make([]float64, len(L))
  for i, l := range L {
    (*A)[i], _ = strconv.ParseFloat(l, 64)
  }
}
func Write(s interface{}) {
  fmt.Fprintln(writer, s)
}
func WriteIntVec(A []int) {
  S := make([]string, len(A))
  for i, a := range A {
    S[i] = strconv.Itoa(a)
  }
  Write(strings.Join(S, " "))
}
func Output() {
  _ = writer.Flush()
}
func MaxInt(A ...int) int {
  max := A[0]
  for _, a := range A {
    if max < a { max = a }
  }
  return max
}
func MinInt(A ...int) int {
  min := A[0]
  for _, a := range A {
    if a < min { min = a }
  }
  return min
}
func SumInt(A ...int) int {
  sum := 0
  for _, a := range A {
    sum += a
  }
  return sum
}
func AbsInt(x int) int {
  if x < 0 { x = -x }
  return x
}
func Ceil(x, y int) int {
  a := x / y
  if x % y > 0 { a++ }
  return a
}