Problem:
Score : 500 points

Problem Statement

There are N cities numbered 1 to N, connected by M railroads.

You are now at City 1, with 10^{100} gold coins and S silver coins in your pocket.

The i-th railroad connects City U_i and City V_i bidirectionally, and a one-way trip costs A_i silver coins and takes B_i minutes.
You cannot use gold coins to pay the fare.

There is an exchange counter in each city. At the exchange counter in City i, you can get C_i silver coins for 1 gold coin.
The transaction takes D_i minutes for each gold coin you give.
You can exchange any number of gold coins at each exchange counter.

For each t=2, ..., N, find the minimum time needed to travel from City 1 to City t. You can ignore the time spent waiting for trains.

Constraints

2 \leq N \leq 50

N-1 \leq M \leq 100

0 \leq S \leq 10^9

1 \leq A_i \leq 50

1 \leq B_i,C_i,D_i \leq 10^9

1 \leq U_i < V_i \leq N

There is no pair i, j(i \neq j) such that (U_i,V_i)=(U_j,V_j).

Each city t=2,...,N can be reached from City 1 with some number of railroads.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N M S
U_1 V_1 A_1 B_1
:
U_M V_M A_M B_M
C_1 D_1
:
C_N D_N

Output

For each t=2, ..., N in this order, print a line containing the minimum time needed to travel from City 1 to City t.

Sample Input 1

3 2 1
1 2 1 2
1 3 2 4
1 11
1 2
2 5

Sample Output 1

2
14

The railway network in this input is shown in the figure below.

In this figure, each city is labeled as follows:

The first line: the ID number i of the city (i for City i)

The second line: C_i / D_i

Similarly, each railroad is labeled as follows:

The first line: the ID number i of the railroad (i for the i-th railroad in input)

The second line: A_i / B_i

You can travel from City 1 to City 2 in 2 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

You can travel from City 1 to City 3 in 14 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 2 minutes.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 6 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Sample Input 2

4 4 1
1 2 1 5
1 3 4 4
2 4 2 2
3 4 1 1
3 1
3 1
5 2
6 4

Sample Output 2

5
5
7

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 4 in 7 minutes, as follows:

At the exchange counter in City 1, exchange 2 gold coins for 6 silver coins in 2 minutes.

Use the 2-nd railroad to move from City 1 to City 3 in 4 minutes.

Use the 4-th railroad to move from City 3 to City 4 in 1 minutes.

Sample Input 3

6 5 1
1 2 1 1
1 3 2 1
2 4 5 1
3 5 11 1
1 6 50 1
1 10000
1 3000
1 700
1 100
1 1
100 1

Sample Output 3

1
9003
14606
16510
16576

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 6 in 16576 minutes, as follows:

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

At the exchange counter in City 2, exchange 3 gold coins for 3 silver coins in 9000 minutes.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

At the exchange counter in City 3, exchange 8 gold coins for 8 silver coins in 5600 minutes.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Use the 3-rd railroad to move from City 2 to City 4 in 1 minute.

At the exchange counter in City 4, exchange 19 gold coins for 19 silver coins in 1900 minutes.

Use the 3-rd railroad to move from City 4 to City 2 in 1 minute.

Use the 1-st railroad to move from City 2 to City 1 in 1 minute.

Use the 2-nd railroad to move from City 1 to City 3 in 1 minute.

Use the 4-th railroad to move from City 3 to City 5 in 1 minute.

At the exchange counter in City 5, exchange 63 gold coins for 63 silver coins in 63 minutes.

Use the 4-th railroad to move from City 5 to City 3 in 1 minute.

Use the 2-nd railroad to move from City 3 to City 1 in 1 minute.

Use the 5-th railroad to move from City 1 to City 6 in 1 minute.

Sample Input 4

4 6 1000000000
1 2 50 1
1 3 50 5
1 4 50 7
2 3 50 2
2 4 50 4
3 4 50 3
10 2
4 4
5 5
7 7

Sample Output 4

1
3
5

The railway network in this input is shown in the figure below:

Sample Input 5

2 1 0
1 2 1 1
1 1000000000
1 1

Sample Output 5

1000000001

The railway network in this input is shown in the figure below:

You can travel from City 1 to City 2 in 1000000001 minutes, as follows:

At the exchange counter in City 1, exchange 1 gold coin for 1 silver coin in 1000000000 minutes.

Use the 1-st railroad to move from City 1 to City 2 in 1 minute.

Solution in OCaml (4.10.0) (judge verdict: Accepted)

module WeightedDirectedGraph
  (Vertex : sig
    type t
    val compare : t -> t -> int
  end)
  (Weight : sig
    type t
    val zero : t
    val ( + ) : t -> t -> t
    val compare : t -> t -> int
  end) :
sig
  val dijkstra :
    (* 隣接リスト *)
    (Vertex.t -> (Vertex.t * Weight.t) list) ->
    (* 始点 *)
    Vertex.t ->
    (* 始点から辿り着けなければNoneを返す *)
    (Vertex.t -> Weight.t option)
end =
struct
  module WMap = Map.Make (Weight)
  module VMap = Map.Make (Vertex)

  let dijkstra e s =
    (* 始点sからの距離 *)
    (* d に入っていない頂点への距離は無限大とみなす *)
    let d = ref @@ VMap.singleton s Weight.zero in
    (* 優先度付きキュー *)
    let q = ref @@ WMap.singleton Weight.zero [s] in
    (* ダイクストラ法のメインループ *)
    let rec dijkstra_aux t =
      match VMap.find_opt t !d, WMap.min_binding_opt !q with
      (* もう既に全ての頂点までの距離が分かっている *)
      | ans, None -> ans
      (* 既に終点までの距離が分かっているので返す *)
      | Some x as ans, Some (w, _) when Weight.compare x w <= 0 -> ans
      (* 終点までの距離が分かっていないので，ダイクストラ法を続行 *)
      | _, Some (w, us) ->
          q := WMap.remove w !q;
          Fun.flip List.iter us (fun u ->
            if 0 <= Weight.compare (VMap.find u !d) w then
              (* 未だ頂点uを訪れていない *)
              Fun.flip List.iter (e u) (fun (v, c) ->
                let open Weight in
                match VMap.find_opt v !d with
                | Some d when Weight.compare d (w + c) <= 0 -> ()
                | _ ->
                    d := VMap.add v (w + c) !d;
                    q := WMap.add (w + c) (v :: try WMap.find (w + c) !q with Not_found -> []) !q));
          dijkstra_aux t in dijkstra_aux
end

module G = WeightedDirectedGraph
  (struct
    type t = int * int
    let compare = compare
  end)
  (struct
    type t = int
    let zero = 0
    let ( + ) = ( + )
    let compare = compare
  end)

let () = Scanf.scanf "%d %d %d\n" @@ fun n m s ->
  let es = Array.make_matrix n 2501 [] in
  for i = 0 to m - 1 do
    Scanf.scanf "%d %d %d %d\n" @@ fun u v a b ->
      for i = a to 2500 do
        es.(u - 1).(i) <- ((v - 1, i - a), b) :: es.(u - 1).(i);
        es.(v - 1).(i) <- ((u - 1, i - a), b) :: es.(v - 1).(i);
      done
  done;
  for v = 0 to n - 1 do
    Scanf.scanf "%d %d\n" @@ fun c d ->
      for j = 0 to 2500 do
        es.(v).(j) <- ((v, min 2500 (j + c)), d) :: es.(v).(j)
      done
  done;
  let d = Fun.flip G.dijkstra (0, min 2500 s) @@ fun (v, c) -> es.(v).(c) in
  for v = 1 to n - 1 do
    Printf.printf "%d\n" @@
    List.fold_left min max_int @@
    List.init 2501 @@ fun i ->
      Option.value ~default:max_int @@ d (v, i)
  done