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: Time Limit Exceeded)

open Batteries
module EnumL = Enum.Labels
module ListL = List.Labels
module ArrayL = Array.Labels

let dbg0 x = Printf.eprintf "[debug]%s\n" @@ dump x
let dbg1 x = dbg0 x; x

let id = identity

let (++) n m = List.range n `To m
let (++-) n m = if n >= m then List.range n `Downto m else []
let (++^) n m = if n < m then List.range n `To (pred m) else []

module ModOp = struct
  let prime = 1_000_000_007
  let (-) a b = (a + prime - b) mod prime
  let (+) a b = (a + b) mod prime
  let ( * ) a b = (a * b) mod prime

  let rec pow x n =
    if n = 0 then 1
    else if n mod 2 = 0 then pow (x*x) (n/2)
    else x * pow (x*x) (n/2)
end

let scan fmt = Scanf.sscanf (read_line ()) fmt

let scan_lines n fmt f =
  if n = 0 then []
  else List.map (fun _ -> scan fmt f) (1 ++ n)

let scan_list ?sep cnv =
  let line = read_line () in
  (match sep with
   | None -> List.map String.of_char @@ String.to_list line
   | Some sep -> String.split_on_char sep line)
  |> List.map cnv

let scan_array ?sep cnv = Array.of_list @@ scan_list ?sep cnv

let scan_matrix n m e ?sep conv =
  let arr = Array.make_matrix n m e in
  ListL.iter (0 ++^ n)
    ~f:(fun i -> arr.(i) <- Array.of_list @@ scan_list ?sep conv);
  arr

let atoi c = Char.code c - Char.code '0'

let rec powerset = function
  | [] -> [[]]
  | hd::tl ->
    let pws =  powerset tl in
    pws @ ListL.map pws ~f:(fun pw -> hd::pw)

let permutations l =
  let rec interleave x = function
    | [] -> [[x]]
    | (hd::tl) as l ->
      (x::l) :: (interleave x tl |> ListL.map ~f:(fun l -> hd::l)) in
  let rec aux = function
    | [] -> [[]]
    | a::rest ->
      aux rest |> ListL.map ~f:(interleave a) |> List.concat in
  aux l

let intersection l =
  EnumL.filter ~f:(fun x -> EnumL.exists ~f:((=) x) @@ Enum.clone l)

let zip l m =
  let n = min (List.length l) (List.length m) in
  List.combine (List.take n l) (List.take n m)

let lower_bound n f =
  let rec aux l u =
    if u - l > 1 then
      let m = (l + u) / 2 in
      if f m then aux l m
      else aux m u
    else u in aux (-1) n

let between n x m = n <= x && x < m

let (n,m,s) = scan "%d %d %d" Tuple3.make
let ls = scan_lines m "%d %d %d %d" Tuple4.make
let ms = Array.of_list @@ scan_lines n "%d %d" Tuple2.make

let num = 2500

let () =
  let mat = Array.make n [] in
  ListL.iter ls ~f:(fun (u,v,a,b) ->
      mat.(pred u) <- (pred v, a, b)::mat.(pred u);
      mat.(pred v) <- (pred u, a, b)::mat.(pred v));
  let dp = Array.make_matrix n (succ num) Int.max_num in
  let rec aux h =
    if Heap.size h = 0 then ()
    else
      let (dist, u, s) = Heap.find_min h in
      if dist > dp.(u).(s) then aux @@ Heap.del_min h
      else
        (let (c,d) = ms.(u) in
         let h = Heap.del_min h in
         let h = if s + c <= num && d + dist < dp.(u).(s+c) then
             Heap.add (d+dist, u, s+c) h
           else h in
         aux @@
         ListL.fold_left mat.(u) ~init:h ~f:(fun h (v,a,b) ->
             if s - a >= 0 then
               let newdist = dist + b in
               if newdist < dp.(v).(s-a) then
                 (dp.(v).(s-a) <- newdist;
                  Heap.add (newdist, v, s-a) h)
               else h
             else h))
  in
  dp.(0).(min s 2500) <- 0;
  aux (Heap.empty |> Heap.add (0, 0, min 2500 s));
  (* Array.print ~sep:"\n" (Array.print Int.print) stderr dp; *)
  ListL.iter (1++^n) ~f:(fun i ->
      Printf.printf "%d\n" @@ Array.min dp.(i))