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 Go (1.6) (judge verdict: Time Limit Exceeded)

package main

import (
	"bufio"
	"container/heap"
	"fmt"
	"math"
	"os"
	"strconv"
)

func getScanner(fp *os.File) *bufio.Scanner {
	scanner := bufio.NewScanner(fp)
	scanner.Split(bufio.ScanWords)
	scanner.Buffer(make([]byte, 500001), 500000)
	return scanner
}

func getNextString(scanner *bufio.Scanner) string {
	scanner.Scan()
	return scanner.Text()
}

func getNextInt(scanner *bufio.Scanner) int {
	i, _ := strconv.Atoi(getNextString(scanner))
	return i
}

func getNextInt64(scanner *bufio.Scanner) int64 {
	i, _ := strconv.ParseInt(getNextString(scanner), 10, 64)
	return i
}

func getNextUint64(scanner *bufio.Scanner) uint64 {
	i, _ := strconv.ParseUint(getNextString(scanner), 10, 64)
	return i
}

func getNextFloat64(scanner *bufio.Scanner) float64 {
	i, _ := strconv.ParseFloat(getNextString(scanner), 64)
	return i
}

func main() {
	fp := os.Stdin
	wfp := os.Stdout
	cnt := 0

	if os.Getenv("MASPY") == "ますピ" {
		fp, _ = os.Open(os.Getenv("BEET_THE_HARMONY_OF_PERFECT"))
		cnt = 1
	}
	if os.Getenv("MASPYPY") == "ますピッピ" {
		wfp, _ = os.Create(os.Getenv("NGTKANA_IS_GENIUS10"))
	}

	scanner := getScanner(fp)
	writer := bufio.NewWriter(wfp)
	solve(scanner, writer)
	for i := 0; i < cnt; i++ {
		fmt.Fprintln(writer, "-----------------------------------")
		solve(scanner, writer)
	}
	writer.Flush()
}

func solve(scanner *bufio.Scanner, writer *bufio.Writer) {
	n := getNextInt(scanner)
	m := getNextInt(scanner)
	s := getNextInt(scanner) - 1
	t := getNextInt(scanner) - 1
	ans := make([]int64, n+1)
	g := make([][]Edge, n)
	rng58 := Kenkoooo{}
	rng58.init(n)
	ysp48 := Kenkoooo{}
	ysp48.init(n)
	for i := 0; i < m; i++ {
		u := getNextInt(scanner) - 1
		v := getNextInt(scanner) - 1
		a := getNextInt64(scanner)
		b := getNextInt64(scanner)
		g[u] = append(g[u], Edge{
			to: v,
			id: i,
			a:  a,
			b:  b,
		})
		g[v] = append(g[v], Edge{
			to: u,
			id: i,
			a:  a,
			b:  b,
		})
	}
	rng58.daike(s, 0, g)
	ysp48.daike(t, 1, g)
	ans[n] = math.MaxInt64
	for i := n - 1; i >= 0; i-- {
		sum := rng58.cost[i] + ysp48.cost[i]
		ans[i] = ans[i+1]
		if sum < ans[i+1] {
			ans[i] = sum
		}
	}

	for i := 0; i < n; i++ {
		fmt.Fprintln(writer, int64(1e15)-ans[i])
	}
}

type Kenkoooo struct {
	cost []int64
	hp   *HP
}

func (k *Kenkoooo) init(n int) {
	k.cost = make([]int64, n)
	for i := 0; i < n; i++ {
		k.cost[i] = math.MaxInt64
	}
	k.hp = &HP{}
	heap.Init(k.hp)
}

func (k *Kenkoooo) daike(s, snuke int, g [][]Edge) {
	heap.Push(k.hp, Cost{
		i:   s,
		sum: 0,
	})
	for k.hp.Len() > 0 {
		p := heap.Pop(k.hp).(Cost)
		if k.cost[p.i] < p.sum {
			continue
		}
		k.cost[p.i] = p.sum
		for _, e := range g[p.i] {
			if snuke == 0 {
				heap.Push(k.hp, Cost{
					i:   e.to,
					sum: p.sum + e.a,
				})
				continue
			}
			heap.Push(k.hp, Cost{
				i:   e.to,
				sum: p.sum + e.b,
			})
		}
	}
}

type Cost struct {
	i   int
	sum int64
}
type HP []Cost

func (h HP) Len() int           { return len(h) }
func (h HP) Less(i, j int) bool { return h[i].sum < h[j].sum }
func (h HP) Swap(i, j int)      { h[i], h[j] = h[j], h[i] }
func (h *HP) Push(x interface{}) {
	*h = append(*h, x.(Cost))
}
func (h *HP) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

// Edge ...
type Edge struct {
	to, id int
	a, b   int64
}