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"
)

var sc *bufio.Scanner

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

const INF = 1000000000000000

type dest struct {
	num  int
	cost int
}

type heapDest []dest

func (h heapDest) Len() int {
	return len(h)
}

func (h heapDest) Less(i, j int) bool {
	return h[i].cost < h[j].cost
}

func (h heapDest) Swap(i, j int) {
	h[i], h[j] = h[j], h[i]
}

func (h *heapDest) Push(x interface{}) {
	*h = append(*h, x.(dest))
}

func (h *heapDest) Pop() interface{} {
	old := *h
	n := len(old)
	x := old[n-1]
	*h = old[0 : n-1]
	return x
}

func dijkstra(n int, s int, adj [][]dest) []int {
	ret := make([]int, n)
	for i := range ret {
		if i != s {
			ret[i] = INF
		}
	}
	h := &heapDest{dest{s, 0}}
	heap.Init(h)
	for len(*h) > 0 {
		d1 := heap.Pop(h).(dest)
		if d1.cost > ret[d1.num] {
			continue
		}
		for _, d2 := range adj[d1.num] {
			if ret[d1.num]+d2.cost < ret[d2.num] {
				ret[d2.num] = ret[d1.num] + d2.cost
				heap.Push(h, dest{d2.num, ret[d1.num] + d2.cost})
			}
		}
	}
	return ret
}

func main() {
	sc = bufio.NewScanner(os.Stdin)
	sc.Buffer(make([]byte, 0), 1000000001*3)
	sc.Split(bufio.ScanWords)
	n, m, s, t := nextInt(), nextInt(), nextInt()-1, nextInt()-1

	adjA, adjB := make([][]dest, n), make([][]dest, n)
	for i := 0; i < m; i++ {
		u, v, a, b := nextInt()-1, nextInt()-1, nextInt(), nextInt()
		adjA[u] = append(adjA[u], dest{v, a})
		adjA[v] = append(adjA[v], dest{u, a})
		adjB[u] = append(adjB[u], dest{v, b})
		adjB[v] = append(adjB[v], dest{u, b})
	}
	costA, costB := dijkstra(n, s, adjA), dijkstra(n, t, adjB)

	ans := make([]int, n+1)
	for i := n - 1; i >= 0; i-- {
		remain := INF - (costA[i] + costB[i])
		switch {
		case remain < ans[i+1]:
			ans[i] = ans[i+1]
		default:
			ans[i] = remain
		}
	}
	for i := 0; i < n; i++ {
		fmt.Println(ans[i])
	}
}