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 PyPy3 (2.4.0) (judge verdict: Accepted)

import sys,queue,math,copy,itertools,bisect,collections
from heapq import *

def main():
    sys.setrecursionlimit(10**7)
    INF = 10**18
    MOD = 10**9 + 7
    LI = lambda : [int(x) for x in sys.stdin.readline().split()]
    NI = lambda : int(sys.stdin.readline())
    SI = lambda : sys.stdin.readline().rstrip()

    n,m,s,t = LI()
    y_root = [[] for _ in range(n)]
    s_root = [[] for _ in range(n)]
    y_node = [INF] * n
    s_node = [INF] * n
    for _ in range(m):
        u,v,a,b = LI()
        y_root[u-1].append((v-1,a))
        y_root[v-1].append((u-1,a))
        s_root[u-1].append((v-1,b))
        s_root[v-1].append((u-1,b))

    def dist(node,root,r):
        q = [(0,r)]
        heapify(q)
        node[r] = 0
        while q:
            d,u = heappop(q)
            if node[u] < d: continue
            for v,e in root[u]:
                if d+e < node[v]:
                    node[v] = d+e
                    heappush(q,(d+e,v))

    dist(y_node,y_root,s-1)
    dist(s_node,s_root,t-1)

    t_node = [(y_node[i] + s_node[i],i) for i in range(n)]
    t_node.sort()

    p = 0
    for i in range(n):
        while t_node[p][1] < i:
            p += 1
        print(10**15 - t_node[p][0])


if __name__ == '__main__':
    main()