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 (
	"container/heap"
	"fmt"
	//"os"
)

type PriorityQueue struct {
	data [][2]int
	len  int
}

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

func (pq *PriorityQueue) Less(i, j int) bool {
	return pq.data[i][0] < pq.data[j][0]
}

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

func (pq *PriorityQueue) Push(x interface{}) {
	pq.data = append(pq.data, x.([2]int))
	pq.len++
}

func (pq *PriorityQueue) Pop() interface{} {
	ret := pq.data[pq.len-1]
	pq.data = pq.data[0 : pq.len-1]
	pq.len--
	return ret
}

func dijk(m [][]int, l int, s int) []int {
	q := PriorityQueue{data: nil, len: 0}
	dist := make([]int, l)
	for i := 0; i < l; i++ {
		dist[i] = 1145141919
	}
	dist[s] = 0
	heap.Init(&q)
	heap.Push(&q, [2]int{0, s})
	for q.Len() != 0 {
		tmp := heap.Pop(&q).([2]int)
		prevcost, src := tmp[0], tmp[1]
		if dist[src] < prevcost {
			continue
		}
		for dest := 0; dest < l; dest++ {
			cost := m[src][dest]
			if dist[dest] > dist[src]+cost {
				dist[dest] = dist[src] + cost
				heap.Push(&q, [2]int{dist[dest], dest})
			}
		}
	}
	return dist
}

func main() {
	var n, m, s, t int
	fmt.Scanf("%d%d%d%d", &n, &m, &s, &t)
	s--
	t--
	ma, mb := make([][]int, n), make([][]int, n)
	for i := 0; i < n; i++ {
		ma[i], mb[i] = make([]int, n), make([]int, n)
		for j := 0; j < n; j++ {
			ma[i][j], mb[i][j] = 1145141919, 1145141919
		}
	}
	for i := 0; i < m; i++ {
		var u, v, a, b int
		fmt.Scanf("%d%d%d%d", &u, &v, &a, &b)
		u--
		v--
		ma[u][v], ma[v][u], mb[u][v], mb[v][u] = a, a, b, b
	}
	cost := make([]int, n)
	acost, bcost := dijk(ma, n, s), dijk(mb, n, t)
	for i := 0; i < n; i++ {
		cost[i] = acost[i] + bcost[i]
	}
	//fmt.Fprintf(os.Stderr, "%v\n", cost)
	mincost := 1145141919
	for i := n - 1; i >= 0; i-- {
		if cost[i] < mincost {
			mincost = cost[i]
		}
		cost[i] = mincost
	}
	for i := 0; i < n; i++ {
		fmt.Printf("%d\n", 1000000000000000-cost[i])
	}

	return
}