Problem:
Score : 500 points

Problem Statement

There is a directed graph with N vertices numbered 1 to N and M edges.
The i-th edge is directed from Vertex A_i to Vertex B_i, and there are C_i coins placed along that edge.
Additionally, there is a button on Vertex N.

We will play a game on this graph.
You start the game on Vertex 1 with zero coins, and head for Vertex N by traversing the edges while collecting coins.
It takes one minute to traverse an edge, and you can collect the coins placed along the edge each time you traverse it.
As usual in games, even if you traverse an edge once and collect the coins, the same number of coins will reappear next time you traverse that edge, which you can collect again.

When you reach Vertex N, you can end the game by pressing the button. (You can also choose to leave Vertex N without pressing the button and continue traveling.)
However, when you end the game, you will be asked to pay T \times P coins, where T is the number of minutes elapsed since the start of the game. If you have less than T \times P coins, you will have to pay all of your coins instead.

Your score will be the number of coins you have after this payment.
Determine if there exists a maximum value of the score that can be obtained. If the answer is yes, find that maximum value.

Constraints

2 \leq N \leq 2500

1 \leq M \leq 5000

1 \leq A_i, B_i \leq N

1 \leq C_i \leq 10^5

0 \leq P \leq 10^5

All values in input are integers.

Vertex N can be reached from Vertex 1.

Input

Input is given from Standard Input in the following format:

N M P
A_1 B_1 C_1
:
A_M B_M C_M

Output

If there exists a maximum value of the score that can be obtained, print that maximum value; otherwise, print -1.

Sample Input 1

3 3 10
1 2 20
2 3 30
1 3 45

Sample Output 1

35

There are two ways to travel from Vertex 1 to Vertex 3:

Vertex 1 \rightarrow 2 \rightarrow 3: You collect 20 + 30 = 50 coins on the way. After two minutes from the start of the game, you press the button, pay 2 \times 10 = 20 coins, and you have 50 - 20 = 30 coins left.

Vertex 1 \rightarrow 2: You collect 45 coins on the way. After one minute from the start of the game, you press the button, pay 1 \times 10 = 10 coins, and you have 45 - 10 = 35 coins left.

Thus, the maximum score that can be obtained is 35.

Sample Input 2

2 2 10
1 2 100
2 2 100

Sample Output 2

-1

The edge extending from Vertex 1 takes you to Vertex 2. If you then traverse the edge extending from Vertex 2 to itself t times and press the button, your score will be 90 + 90t. Thus, you can infinitely increase your score, which means there is no maximum value of the score that can be obtained.

Sample Input 3

4 5 10
1 2 1
1 4 1
3 4 1
2 2 100
3 3 100

Sample Output 3

0

There is no way to travel from Vertex 1 to Vertex 4 other than traversing the edge leading from Vertex 1 to Vertex 4 directly. You will pick up one coin along this edge, but after being asked to paying 10 coins, your score will be 0.

Note that you can collect an infinite number of coins if you traverse the edge leading from Vertex 1 to Vertex 2, but this is pointless since you can no longer reach Vertex 4 and end the game.

Solution in Fortran (gfortran v4.8.4) (judge verdict: Accepted)

module mod_graph
  implicit none
  integer(8), parameter :: inf = 1000000000000000000_8

  type item
    integer :: id = 0
    integer(8) :: wt = 0_8
  end type

  type priorityqueue
    integer :: num = 0
    type(item), pointer :: heap(:) => null()
  contains
    !final :: finalizei
  end type

  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
    procedure :: bellman_ford => bellman_ford
    !final :: finalizee
  end type

  type graph
    type(arraylist), pointer :: egs(:) => null()
    logical, pointer :: used(:) => null()
  contains
    procedure :: add => add
    procedure :: dijkstra => dijkstra
    procedure :: ford_fulkerson => ford_fulkerson
    !final :: finalize
  end type

  interface graph
    module procedure :: newg
  end interface graph

contains

  function newi(id,wt) result(ret)
    integer, intent(in) :: id
    integer(8), intent(in) :: wt
    type(item) :: ret
    ret%id = id
    ret%wt = wt
  end

  subroutine swapi(a,b)
    type(item), intent(inout) :: a, b
    type(item) :: c
    c = a
    a = b
    b = c
  end

  subroutine appendi(a)
    type(item), pointer, intent(inout) :: a(:)
    integer :: n
    type(item), allocatable :: tmp(:)
    n = size(a)
    allocate(tmp(n))
    tmp = a
    deallocate(a)
    allocate(a(2*n))
    a(1:n) = tmp
    deallocate(tmp)
  end

  function lessi(a,b) result(ret)
    type(item), intent(in) :: a, b
    logical :: ret
    ret = a%wt < b%wt
  end

  subroutine finalizei(pq)
    type(priorityqueue), intent(inout) :: pq
    if (associated(pq%heap)) deallocate(pq%heap)
    pq%num = 0
  end

  subroutine offer(pq,it)
    type(priorityqueue), intent(inout) :: pq
    type(item), intent(in) :: it
    integer :: i, j
    if (.not.associated(pq%heap)) allocate(pq%heap(1))
    if (pq%num == size(pq%heap)) call appendi(pq%heap)
    pq%num = pq%num+1
    pq%heap(pq%num) = it
    i = pq%num
    do while (i > 1)
      j = i/2
      if (lessi(pq%heap(i),pq%heap(j))) call swapi(pq%heap(i),pq%heap(j))
      i = j
    end do
  end

  function poll(pq) result(ret)
    type(priorityqueue), intent(inout) :: pq
    type(item) :: ret
    integer :: n, i, j
    n = pq%num
    ret = pq%heap(1)
    pq%heap(1) = pq%heap(n)
    pq%num = pq%num-1
    i = 1
    do while (2*i < n)
      j = 2*i
      if (j+1 < n .and. lessi(pq%heap(j+1),pq%heap(j))) j = j+1
      if (lessi(pq%heap(j),pq%heap(i))) call swapi(pq%heap(j),pq%heap(i))
      i = j
    end do
  end

  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 finalizee(list)
    type(arraylist), intent(inout) :: list
    if (associated(list%arr)) deallocate(list%arr)
    list%num = 0
  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

  subroutine finalize(g)
    type(graph), intent(inout) :: g
    integer :: i
    if (associated(g%used)) deallocate(g%used)
    if (.not.associated(g%egs)) return
    do i = 1, size(g%egs)
      call finalizee(g%egs(i))
    end do
    deallocate(g%egs)
  end

  function dijkstra(g,s) result(ret)
    class(graph), intent(in) :: g
    integer, intent(in) :: s
    integer(8) :: ret(size(g%egs))
    type(priorityqueue) :: pq
    type(item) :: it
    type(edge) :: e
    integer :: i
    ret = inf
    ret(s) = 0_8
    call offer(pq,newi(s,ret(s)))
    do while (pq%num > 0)
      it = poll(pq)
      if (it%wt > ret(it%id)) cycle
      do i = 1, g%egs(it%id)%num
        e = g%egs(it%id)%arr(i)
        if (ret(e%to) > ret(it%id)+e%wt) then
          ret(e%to) = ret(it%id)+e%wt
          call offer(pq,newi(e%to,ret(e%to)))
        end if
      end do
    end do
    call finalizei(pq)
  end

  recursive subroutine dfs_bf(g,u)
    type(graph), intent(inout) :: g
    integer, intent(in) :: u
    integer :: i, v
    g%used(u) = .true.
    do i = 1, g%egs(u)%num
      v = g%egs(u)%arr(i)%to
      if (g%used(v)) cycle
      call dfs_bf(g,v)
    end do
  end

  function bellman_ford(es,s,t) result(ret)
    class(arraylist), intent(in) :: es
    integer, intent(in) :: s, t
    integer(8) :: ret
    integer :: n, i, fm, to, step
    integer(8) :: wt
    logical :: updated
    integer(8), allocatable :: tmp(:)
    type(graph) :: gs, gt
    type(arraylist) :: reach
    n = 0
    do i = 1, es%num
      n = max(n,es%arr(i)%fm,es%arr(i)%to)
    end do
    gs = newg(n)
    gt = newg(n)
    allocate(gs%used(n),gt%used(n))
    gs%used = .false.
    gt%used = .false.
    do i = 1, es%num
      fm = es%arr(i)%fm
      to = es%arr(i)%to
      wt = es%arr(i)%wt
      call add(gs,fm,to,wt)
      call add(gt,to,fm,wt)
    end do
    call dfs_bf(gs,s)
    call dfs_bf(gt,t)
    n = 0
    do i = 1, es%num
      fm = es%arr(i)%fm
      to = es%arr(i)%to
      if (gs%used(fm) .and. gt%used(fm) .and. gs%used(to) .and. gt%used(to)) then
        call adde(reach,es%arr(i))
        n = max(n,fm,to)
      end if
    end do
    call finalize(gs)
    call finalize(gt)
    allocate(tmp(n))
    tmp = inf
    tmp(s) = 0_8
    step = 0
    updated = .true.
    ret = inf
    do while (updated)
      updated = .false.
      do i = 1, reach%num
        fm = reach%arr(i)%fm
        to = reach%arr(i)%to
        wt = reach%arr(i)%wt
        if (tmp(to) > tmp(fm)+wt) then
          tmp(to) = tmp(fm)+wt
          updated = .true.
        end if
      end do
      step = step+1
      if (step > n) then
        ret = -inf
        exit
      end if
    end do
    if (ret == inf) ret = tmp(t)
    deallocate(tmp)
    call finalizee(reach)
  end

  recursive function dfs_ff(g,u,t,f) result(ret)
    class(graph), intent(inout) :: g
    integer, intent(in) :: u, t, f
    integer :: ret
    integer :: i, v, cp, rv
    ret = f
    if (u == t) return
    g%used(u) = .true.
    do i = 1, g%egs(u)%num
      v = g%egs(u)%arr(i)%to
      cp = g%egs(u)%arr(i)%cp
      if (g%used(v) .or. cp <= 0) cycle
      ret = dfs_ff(g,v,t,min(f,cp))
      rv = g%egs(u)%arr(i)%rv
      if (ret > 0) then
        g%egs(u)%arr(i)%cp = g%egs(u)%arr(i)%cp-ret
        g%egs(v)%arr(rv)%cp = g%egs(v)%arr(rv)%cp+ret
        return
      end if
    end do
    ret = 0
  end

  function ford_fulkerson(g,s,t) result(ret)
    class(graph), intent(inout) :: g
    integer, intent(in) :: s, t
    integer :: ret
    integer :: f
    ret = 0
    if (associated(g%used) .and. size(g%used) /= size(g%egs)) deallocate(g%used)
    if (.not.associated(g%used)) allocate(g%used(size(g%egs)))
    do
      g%used = .false.
      f = dfs_ff(g,s,t,1000000000)
      if (f == 0) then
        deallocate(g%used)
        return
      end if
      ret = ret+f
    end do
  end

end module mod_graph

program coins_respawn
  use mod_graph
  implicit none
  integer :: n, m, p, a, b, c, i
  type(arraylist) :: es
  integer(8) :: ans = 0_8
  read(*,*) n, m, p
  do i = 1, m
    read(*,*) a, b, c
    call es%add(newe(b,int(p-c,8),a))
  end do
  ans = es%bellman_ford(1,n)
  if (ans == -inf) then
    ans = -1_8
  else if (ans > 0_8) then
    ans = 0_8
  else
    ans = -ans
  end if
  write(*,'(i0)') ans
end program coins_respawn