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 Fortran (GNU Fortran 9.2.1) (judge verdict: Wrong Answer)

program knapsack_queries_on_a_tree
  implicit none
  integer :: n, q, u(100000,3) = 0, p(0:18) = -1, m, r, h, i, j, k
  integer, dimension(262144) :: v = 0, w = 0
  integer(8) :: dp(0:18,100000) = 0, ans(100000) = 0
  read(*,*) n
  do i = 1, n
    read(*,*) v(i), w(i)
  end do
  read(*,*) q
  do i = 1, q
    read(*,*) u(i,1:2)
    u(i,3) = i
  end do
  call merge_sort(u(1:q,1:3),1)
  p(0) = 1
  dp(0,w(1):100000) = v(1)
  do i = 1, q
    m = u(i,1)
    h = headbit(m)
    do j = 1, h
      r = rshift(m,h-j)
      if (p(j) == r) cycle
      p(j) = r
      dp(j,:) = 0
      do k = 100000, w(r), -1
        dp(j,k) = max(dp(j-1,k),dp(j-1,k-w(r))+v(r))
      end do
    end do
    ans(u(i,3)) = dp(h,u(i,2))
  end do
  do i = 1, q
    write(*,'(i0)') ans(i)
  end do
contains
  integer function headbit(n)
    integer, intent(in) :: n
    integer :: i
    do i = 18, 0, -1
      if (btest(n,i)) then
        headbit = i
        return
      end if
    end do
  end
  subroutine merge_sort(a,k)
    integer, intent(inout) :: a(:,:)
    integer, intent(in) :: k
    integer :: n, m, l, i, u
    n = size(a,1)
    m = n
    l = 1
    do while (m > 1)
      do i = 1, m/2
        u = min(2*i*l,n)
        call merger(a(2*(i-1)*l+1:(2*i-1)*l,:),a((2*i-1)*l+1:u,:),k)
      end do
      l = 2*l
      m = (m+1)/2
    end do
  end
  subroutine merger(a1,a2,k)
    integer, intent(inout) :: a1(:,:), a2(:,:)
    integer, intent(in) :: k
    integer :: a(size(a1,1)+size(a2,1),size(a1,2))
    integer :: i1, i2, n1, n2
    i1 = 1
    i2 = 1
    n1 = size(a1,1)
    n2 = size(a2,1)
    do while (i1 <= n1 .and. i2 <= n2)
      if (a1(i1,k) <= a2(i2,k)) then
        a(i1+i2-1,:) = a1(i1,:)
        i1 = i1+1
      else
        a(i1+i2-1,:) = a2(i2,:)
        i2 = i2+1
       end if
    end do
    if (i1 <= n1) then
      a(i1+i2-1:n1+n2,:) = a1(i1:n1,:)
    else
      a(i1+i2-1:n1+n2,:) = a2(i2:n2,:)
    end if
    a1 = a(1:n1,:)
    a2 = a(n1+1:n1+n2,:)
  end
end program knapsack_queries_on_a_tree