Problem:
Score : 500 points

Problem Statement

There are N towns numbered 1 to N and M roads. The i-th road connects Town A_i and Town B_i bidirectionally and has a length of C_i.

Takahashi will travel between these towns by car, passing through these roads. The fuel tank of his car can contain at most L liters of fuel, and one liter of fuel is consumed for each unit distance traveled. When visiting a town while traveling, he can full the tank (or choose not to do so). Travel that results in the tank becoming empty halfway on the road cannot be done.

Process the following Q queries:

The tank is now full. Find the minimum number of times he needs to full his tank while traveling from Town s_i to Town t_i. If Town t_i is unreachable, print -1.

Constraints

All values in input are integers.

2 \leq N \leq 300

0 \leq M \leq \frac{N(N-1)}{2}

1 \leq L \leq 10^9

1 \leq A_i, B_i \leq N

A_i \neq B_i

\left(A_i, B_i\right) \neq \left(A_j, B_j\right) (if i \neq j)

\left(A_i, B_i\right) \neq \left(B_j, A_j\right) (if i \neq j)

1 \leq C_i \leq 10^9

1 \leq Q \leq N\left(N-1\right)

1 \leq s_i, t_i \leq N

s_i \neq t_i

\left(s_i, t_i\right) \neq \left(s_j, t_j\right) (if i \neq j)

Input

Input is given from Standard Input in the following format:

N M L
A_1 B_1 C_1
:
A_M B_M C_M
Q
s_1 t_1
:
s_Q t_Q

Output

Print Q lines.

The i-th line should contain the minimum number of times the tank needs to be fulled while traveling from Town s_i to Town t_i. If Town t_i is unreachable, the line should contain -1 instead.

Sample Input 1

3 2 5
1 2 3
2 3 3
2
3 2
1 3

Sample Output 1

0
1

To travel from Town 3 to Town 2, we can use the second road to reach Town 2 without fueling the tank on the way.

To travel from Town 1 to Town 3, we can first use the first road to get to Town 2, full the tank, and use the second road to reach Town 3.

Sample Input 2

4 0 1
1
2 1

Sample Output 2

-1

There may be no road at all.

Sample Input 3

5 4 4
1 2 2
2 3 2
3 4 3
4 5 2
20
2 1
3 1
4 1
5 1
1 2
3 2
4 2
5 2
1 3
2 3
4 3
5 3
1 4
2 4
3 4
5 4
1 5
2 5
3 5
4 5

Sample Output 3

0
0
1
2
0
0
1
2
0
0
0
1
1
1
0
0
2
2
1
0

Solution in Fortran (gfortran v4.8.4) (judge verdict: Time Limit Exceeded)

module mod_graph
  implicit none
  integer(8), parameter :: inf = 10000000
  type edge
    integer :: fm = 0, to = 0, cp = 0, rv = 0
    integer(8) :: wt = 0_8
  end type
  type arraylist
    integer :: num = 0
    type(edge), pointer :: arr(:) => null()
  contains
    procedure :: add => adde
  end type
  type graph
    type(arraylist), pointer :: egs(:) => null()
    logical, pointer :: used(:) => null()
  contains
    procedure :: add => add
  end type
  interface graph
    module procedure :: newg
  end interface graph
contains
  function newe(to,wt,fm,cp,rv) result(ret)
    integer, intent(in) :: to
    integer(8), intent(in) :: wt
    integer, intent(in), optional :: fm, cp, rv
    type(edge) :: ret
    if (present(fm)) ret%fm = fm
    ret%to = to
    ret%wt = wt
    if (present(cp) .and. present(rv)) then
      ret%cp = cp
      ret%rv = rv
    end if
  end
  subroutine appende(a)
    type(edge), pointer, intent(inout) :: a(:)
    integer :: n
    type(edge), allocatable :: tmp(:)
    n = size(a)
    allocate(tmp(n))
    tmp = a
    deallocate(a)
    allocate(a(2*n))
    a(1:n) = tmp
    deallocate(tmp)
  end
  function lesse(a,b) result(ret)
    type(edge), intent(in) :: a, b
    logical :: ret
    ret = a%wt < b%wt
  end
  subroutine adde(list,e)
    class(arraylist), intent(inout) :: list
    type(edge), intent(in) :: e
    if (.not.associated(list%arr)) allocate(list%arr(1))
    if (list%num == size(list%arr)) call appende(list%arr)
    list%num = list%num+1
    list%arr(list%num) = e
  end
  function newg(n) result(ret)
    integer, intent(in) :: n
    type(graph) :: ret
    allocate(ret%egs(n))
  end
  subroutine add(g,fm,to,wt,cp)
    class(graph), intent(inout) :: g
    integer, intent(in) :: fm, to
    integer, intent(in), optional :: cp
    integer(8) :: wt
    if (present(cp)) then
      call adde(g%egs(fm),newe(to,wt,cp=cp,rv=g%egs(to)%num))
      call adde(g%egs(to),newe(fm,wt,cp=0,rv=g%egs(fm)%num-1))
    else
      call adde(g%egs(fm),newe(to,wt))
    end if
  end
end module mod_graph
program travel_by_car
  use mod_graph
  implicit none
  integer :: n, m, a, b, q, s, t, i, j
  integer(8) :: l, c
  integer :: x(300,300) = inf
  type(graph) :: g
  read(*,*) n, m, l
  g = graph(n)
  do i = 1, m
    read(*,*) a, b, c
    call g%add(a,b,c)
    call g%add(b,a,c)
  end do
  do i = 1, n
    call dfs(i,i,l,0)
  end do
  read(*,*) q
  do i = 1, q
    read(*,*) s, t
    if (x(s,t) == inf) then
      write(*,'(i0)') -1
    else
      write(*,'(i0)') x(s,t)
    end if
  end do
contains
  recursive subroutine dfs(s,u,gas,m)
    integer, intent(in) :: s, u, m
    integer(8), intent(in) :: gas
    integer :: v, i
    integer(8) :: c
    if (x(s,u) < m) return
    x(s,u) = m
    do i = 1, g%egs(u)%num
      v = g%egs(u)%arr(i)%to
      c = g%egs(u)%arr(i)%wt
      if (gas >= c) then
        call dfs(s,v,gas-c,m)
      else if (l >= c) then
        call dfs(s,v,l-c,m+1)
      end if
    end do
  end
end program travel_by_car