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 Julia (1.4.0) (judge verdict: Accepted)

const double = Float64
const int = Int64
const vint = Vector{int}
const vdouble = Vector{double}

parseint(x) = parse(int, x)
function readint()
  moji = readline(stdin)
  moji = split(moji)
  return map(parseint,moji)
end

const NMax = 50
const MMax = 100

struct Edge
  y::int
  cost ::int
  time ::int
end

mutable struct Graph{T}
  nodes::Vector{Vector{T}} 
  degree::Vector{int}
  n::int
  m::int
end

function Graph(N,T)
  Nodes = Vector{Vector{T}}(undef,N)
  for i in 1:N
    Nodes[i] = Vector{T}(undef,0)
  end
  degree = zeros(int,N)
  n = N
  m = 0
  return Graph{T}(Nodes, degree, n,m)

end

struct State
  iti::int
  kane::int
  time::int
end

isless(a::State,b::State) = a.time < b.time

mutable struct PriorityQueue{T}
  tree::Vector{T}
  Nmax::int
  n::int
  priority::Function
end

function PriorityQueue(isless,T)
  tree = Vector{T}(undef,0)
  return PriorityQueue{T}(tree, 0, 0, isless)
end

function push(P::PriorityQueue{T}, a::T) where T
  if P.Nmax == P.n
    P.Nmax += 1
    P.n += 1
    append!(P.tree,[a])
  else
    P.n+=1
    P.tree[P.n] = a
  end
  
  p = P.n
  while p > 1
    up = p÷2
    if P.priority(P.tree[p], P.tree[up]) 
      P.tree[p], P.tree[up] = P.tree[up],P.tree[p]
    end
    p = up
  end
      
end
  
function pop(P::PriorityQueue{T}) where T
  preturn = P.tree[1]
  P.tree[1] = P.tree[P.n]
  P.n -= 1 
  nmax = P.n
  p = 1
  
  while 2p <= nmax
    
    # argmin [p, 2p, 2p+1]
    m = p
    for down in 2p:2p+1
      down > nmax && break
      P.priority(P.tree[down], P.tree[m]) && (m = down)
    end
    
    m == p && break
    
    P.tree[p], P.tree[m] = P.tree[m], P.tree[p]
    p = m
        
  end
  return preturn

end

#import 
is_empty(P::PriorityQueue{T}) where T = P.n == 0

function main()
  N, M, S = readint()
  
  G =Graph(N,Edge)

  Amax = 0
  for i in 1:M
    U,V,A,B = readint()
    append!(G.nodes[U],[Edge(V,A,B)])
    append!(G.nodes[V],[Edge(U,A,B)])
    G.degree[U]+=1
    G.degree[V]+=1
    Amax < A && (Amax = A)
  end

  for i in 1:N
    C,D = readint()
    G.degree[i] += 1
    append!(G.nodes[i], [Edge(i,-C,D)])
  end

  d = Array{int,2}(undef,N,N*Amax+1)
  d .= typemax(int)
  if S > Amax * N 
    S = Amax * N
  end
  d[1, S+1] = 0

  checked = Vector{Bool}(undef,N)
  checked .= false

  Q = PriorityQueue(isless, State)
  push(Q, State(1,S,0))

  jS = S
  jp = 1
  checked[1]=true
  dd = Vector{int}(undef, N)

  while ! is_empty(Q)
    q = pop(Q)
    p, kane, time = q.iti, q.kane, q.time
    
    d[p,kane+1] < time && continue
  
    if !checked[p] # = false
      checked[p] = true
      dd[p] = time
    end

    for E in G.nodes[p]
      kane - E.cost < 0 && continue

      newkane = min(Amax*N, kane-E.cost)
      newtime = E.time + time 

      if d[E.y, newkane+1] > newtime
        d[E.y, newkane+1] = newtime   
        push(Q, State(E.y, newkane, newtime))
      end
    end
    #println(q, p)    
  end
  for i in 2:N
    println(dd[i])
  end
  #println(G)
end

main()