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: Accepted)

package main

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

func main() {
	sc = bufio.NewScanner(os.Stdin)
	sc.Buffer(make([]byte, 0), 1000000001*3)
	sc.Split(bufio.ScanWords)

	var n, m, s, t int
	fmt.Scan(&n, &m, &s, &t)
	s, t = s-1, t-1

	gy := make([][]Edge, n)
	gs := make([][]Edge, n)
	for i := 0; i < n; i++ {
		gy[i] = make([]Edge, 0)
		gs[i] = make([]Edge, 0)
	}

	for i := 0; i < m; i++ {
		u, v, a, b := nextInt()-1, nextInt()-1, nextInt(), nextInt()
		gy[u] = append(gy[u], Edge{to: v, cost: a})
		gy[v] = append(gy[v], Edge{to: u, cost: a})
		gs[u] = append(gs[u], Edge{to: v, cost: b})
		gs[v] = append(gs[v], Edge{to: u, cost: b})
	}

	cy := dijkstra(gy, s)
	cs := dijkstra(gs, t)
	co := make([]int, n)
	for i := 0; i < n; i++ {
		co[i] = cy[i] + cs[i]
	}
	for i := n - 2; i >= 0; i-- {
		co[i] = min(co[i], co[i+1])
	}
	for i := 0; i < n; i++ {
		fmt.Println(1e15 - co[i])
	}
}

func min(a, b int) int {
	if a < b {
		return a
	}
	return b
}

func dijkstra(g [][]Edge, from int) []int {
	n := len(g)
	d := make([]int, n)
	for i := 0; i < n; i++ {
		d[i] = 1e15
	}

	var pq PriorityQueue
	heap.Init(&pq)

	heap.Push(&pq, &Item{pos: from, cost: 0})
	d[from] = 0

	for len(pq) > 0 {
		s := heap.Pop(&pq).(*Item)
		if d[s.pos] < s.cost {
			continue
		}
		for _, e := range g[s.pos] {
			if d[e.to] > s.cost+e.cost {
				d[e.to] = s.cost + e.cost
				heap.Push(&pq, &Item{pos: e.to, cost: d[e.to]})
			}
		}
	}

	return d
}

// Edge .
type Edge struct {
	to, cost int
}

// https://golang.org/pkg/container/heap/

// An Item is something we manage in a priority queue.
type Item struct {
	pos  int // The value of the item; arbitrary.
	cost int // The priority of the item in the queue.
	// The index is needed by update and is maintained by the heap.Interface methods.
	index int // The index of the item in the heap.
}

// A PriorityQueue implements heap.Interface and holds Items.
type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }

func (pq PriorityQueue) Less(i, j int) bool {
	// We want Pop to give us the highest, not lowest, priority so we use greater than here.
	return pq[i].cost < pq[j].cost
}

func (pq PriorityQueue) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

// Push .
func (pq *PriorityQueue) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Item)
	item.index = n
	*pq = append(*pq, item)
}

// Pop .
func (pq *PriorityQueue) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	old[n-1] = nil  // avoid memory leak
	item.index = -1 // for safety
	*pq = old[0 : n-1]
	return item
}

var sc *bufio.Scanner

func nextInt() int {
	sc.Scan()
	i, e := strconv.Atoi(sc.Text())
	if e != nil {
		panic(e)
	}
	return i
}