Problem:
Score : 400 points

Problem Statement

Kenkoooo is planning a trip in Republic of Snuke.
In this country, there are n cities and m trains running.
The cities are numbered 1 through n, and the i-th train connects City u_i and v_i bidirectionally.
Any city can be reached from any city by changing trains.

Two currencies are used in the country: yen and snuuk.
Any train fare can be paid by both yen and snuuk.
The fare of the i-th train is a_i yen if paid in yen, and b_i snuuk if paid in snuuk.

In a city with a money exchange office, you can change 1 yen into 1 snuuk.
However, when you do a money exchange, you have to change all your yen into snuuk.
That is, if Kenkoooo does a money exchange when he has X yen, he will then have X snuuk.
Currently, there is a money exchange office in every city, but the office in City i will shut down in i years and can never be used in and after that year.

Kenkoooo is planning to depart City s with 10^{15} yen in his pocket and head for City t, and change his yen into snuuk in some city while traveling.
It is acceptable to do the exchange in City s or City t.

Kenkoooo would like to have as much snuuk as possible when he reaches City t by making the optimal choices for the route to travel and the city to do the exchange.
For each i=0,...,n-1, find the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i years.
You can assume that the trip finishes within the year.

Constraints

2 \leq n \leq 10^5

1 \leq m \leq 10^5

1 \leq s,t \leq n

s \neq t

1 \leq u_i < v_i \leq n

1 \leq a_i,b_i \leq 10^9

If i\neq j, then u_i \neq u_j or v_i \neq v_j.

Any city can be reached from any city by changing trains.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

n m s t
u_1 v_1 a_1 b_1
:
u_m v_m a_m b_m

Output

Print n lines.
In the i-th line, print the maximum amount of snuuk that Kenkoooo has when he reaches City t if he goes on a trip from City s to City t after i-1 years.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

8 12 3 8
2 8 685087149 857180777
6 7 298270585 209942236
2 4 346080035 234079976
2 5 131857300 22507157
4 8 30723332 173476334
2 6 480845267 448565596
1 4 181424400 548830121
4 5 57429995 195056405
7 8 160277628 479932440
1 6 475692952 203530153
3 5 336869679 160714712
2 7 389775999 199123879

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in Haskell (GHC 7.10.3) (judge verdict: Wrong Answer)

{-# LANGUAGE BangPatterns, OverloadedStrings #-}
import Control.Arrow
import Control.Monad.Fix
import Data.Foldable
import Data.Monoid ((<>))
import qualified Data.Vector as V
import qualified Data.Vector.Mutable as VM
import qualified Data.Vector.Unboxed as VU
import qualified Data.Vector.Unboxed.Mutable as VUM
import qualified Data.ByteString.Char8 as B
import qualified Data.ByteString.Builder as B
import qualified Data.ByteString.Lazy as BL
import Data.ByteString.Internal (accursedUnutterablePerformIO)
import System.IO (stdout)
import Debug.Trace

uread :: VUM.Unbox a => Int -> VUM.IOVector a -> a
uread !i !v = accursedUnutterablePerformIO $ VUM.unsafeRead v i

uwrite :: VUM.Unbox a => Int -> a -> VUM.IOVector a -> VUM.IOVector a
uwrite !i !a !v =
  accursedUnutterablePerformIO $ VUM.unsafeWrite v i a >> return v

uswap :: VUM.Unbox a => Int -> Int -> VUM.IOVector a -> VUM.IOVector a
uswap !i !j !v =
  accursedUnutterablePerformIO $ VUM.unsafeSwap v i j >> return v

uthaw :: VUM.Unbox a => VU.Vector a -> VUM.IOVector a
uthaw !v = accursedUnutterablePerformIO $ VU.unsafeThaw v

ufreeze :: VUM.Unbox a => VUM.IOVector a -> VU.Vector a
ufreeze !v = accursedUnutterablePerformIO $ VU.unsafeFreeze v

data PQueue a = PQueue !Int !(VUM.IOVector a)

new :: VUM.Unbox a => Int -> PQueue a
new !s = PQueue 0 $ accursedUnutterablePerformIO $ VUM.new (s + 1)

push :: (Ord a, Bounded a) => VUM.Unbox a => a -> PQueue a -> PQueue a
push !a !(PQueue k vec) = PQueue (k + 1) vec'
 where
  vec' = up k a $ uwrite k a $ if VUM.length vec <= k
    then accursedUnutterablePerformIO $ VUM.unsafeGrow vec (VUM.length vec * 2)
    else vec

  up !i !cv !vec = (\(_, v) -> v) $ fix
    ( \f (!ci, !vec) ->
      let pi = (ci - 1) `div` 2
          pv = if 0 < ci then uread pi vec else minBound
      in  if 0 < ci && cv < pv
            then let vec' = uswap pi ci vec in vec' `seq` f (pi, vec')
            else (ci, vec)
    )
    (i, vec)

popMin :: (Ord a, Bounded a, VUM.Unbox a) => PQueue a -> Maybe (a, PQueue a)
popMin !(PQueue k vec) | k <= 0    = Nothing
                       | otherwise = Just (uread 0 vec, PQueue (k - 1) vec')
 where
  vec' = let v = uswap (k - 1) 0 vec in v `seq` down 0 (uread 0 vec) v

  down !i !cv !vec = (\(_, v) -> v) $ fix
    ( \f (!ci, !vec) ->
      let li = ci * 2 + 1
          lv = if li < k - 1 then uread li vec else maxBound
          ri = ci * 2 + 2
          rv = if ri < k - 1 then uread ri vec else maxBound
      in  if cv <= lv && cv <= rv
            then (ci, vec)
            else
              let p    = if lv > rv then ri else li
                  vec' = uswap p ci vec
              in  vec' `seq` f (p, vec')
    )
    (i, vec)

popMin' :: (Ord a, Bounded a, VUM.Unbox a) => PQueue a -> (a, PQueue a)
popMin' !pq = maybe (error "size < 0") id $ popMin pq

umodifyV :: Int -> (a -> a) -> VM.IOVector a -> VM.IOVector a
umodifyV !i !f !v = accursedUnutterablePerformIO $ VM.modify v f i >> return v

uthawV :: V.Vector a -> VM.IOVector a
uthawV !v = accursedUnutterablePerformIO $ V.thaw v

ufreezeV :: VM.IOVector a -> V.Vector a
ufreezeV !v = accursedUnutterablePerformIO $ V.freeze v

dijkstra
  :: Int  -- ^ nodes
  -> V.Vector [(Int, Int)]  -- ^ edges from i-th node: (end, cost)
  -> Int  -- ^ start node
  -> Int  -- ^ end node
  -> VU.Vector Int  -- ^ constructed graph
dijkstra !n !edges !s !t = ufreeze $ snd $ fix
  ( \f (!nodes, !costs) -> case popMin nodes of
    Nothing              -> (nodes, costs)
    Just ((c, n), nodes) -> if c <= uread n costs
      then
        let es = edges V.! n
        in  f $ foldl'
              ( \(!ns, !cs) (!e, !ec) ->
                let c' = c + ec
                in  if c' < uread e costs
                      then (push (c', e) ns, uwrite e c' cs)
                      else (ns, cs)
              )
              (nodes, costs)
              es
      else f (nodes, costs)
  )
  (nodes0, costs0)
 where
  costs0 :: VUM.IOVector Int
  costs0 = uwrite s 0 $ uthaw $ VU.replicate n maxBound

  nodes0 :: PQueue (Int, Int)
  nodes0 = push (0, s) $ new 0

solve :: Int -> Int -> Int -> Int -> VU.Vector (Int, Int, Int, Int) -> [Int]
solve !n !m !s !t !trains = map (10 ^ 15 -) $ fst $ foldl'
  (\(!acc, !prev) !i -> let cur = calcCost i `min` prev in (0 : acc, 0))
  ([], 10 ^ 15 :: Int)
  [n - 1, n - 2 .. 0]
 where
  yenGraph = dijkstra
    n
    ( ufreezeV $ VU.foldl'
      (\gr (u, v, a, b) -> umodifyV v ((u, a) :) $ umodifyV u ((v, a) :) $ gr)
      (uthawV $ V.replicate n [])
      trains
    )
    s
    t
  snookGraph = dijkstra
    n
    ( ufreezeV $ VU.foldl'
      (\gr (u, v, a, b) -> umodifyV v ((u, b) :) $ umodifyV u ((v, b) :) $ gr)
      (uthawV $ V.replicate n [])
      trains
    )
    t
    s

  calcCost i = yenGraph VU.! i + snookGraph VU.! i

main = do
  let readInt = fmap (second B.tail) . B.readInt

  (n, m, s, t) <-
    (\vec -> (vec VU.! 0, vec VU.! 1, vec VU.! 2 - 1, vec VU.! 3 - 1))
    .   VU.unfoldrN 4 readInt
    <$> B.getLine
  uvab <-
    VU.replicateM m
    $   (\vec -> (vec VU.! 0 - 1, vec VU.! 1 - 1, vec VU.! 2, vec VU.! 3))
    .   VU.unfoldrN 4 readInt
    <$> B.getLine

  B.hPutBuilder stdout $ foldl' (\acc x -> acc <> B.intDec x <> "\n") "" $ solve
    n
    m
    s
    t
    uvab