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: Time Limit Exceeded)

#### import ####
import sys
import math
from collections import defaultdict

#### 設定 ####
sys.setrecursionlimit(10**7)
def input():
  return sys.stdin.readline()[:-1]

#### 定数 ####
mod = 10**9 + 7

#### 読み込み ####
def I(): return int(input())
def II(): return map(int, input().split())
def III(): return list(map(int, input().split()))
def Line(N):
  read_all = [tuple(map(int, input().split())) for _ in range(N)]
  return map(list,zip(*read_all))

#################

n,m,s,t = II()
u,v,a,b = Line(m)

from heapq import heappop, heappush
def dijkstra(E,s,n):
  inf=float('inf')
  d=[inf for _ in range(n)]
  d[s]=0
  q=[]
  heappush(q,(0,s))
  while q:
    du, u1 = heappop(q)
    if d[u1]<du: continue
    for v1,weight in E.get(u1,[]):
      alt=du+weight
      if d[v1]>alt:
        d[v1]=alt
        heappush(q, (alt,v1))
  return d

p = dict()
p2 = dict()
for i in range(m):
  p[u[i]-1]=p.get(u[i]-1,[])+[(v[i]-1,a[i])]
  p[v[i]-1]=p.get(v[i]-1,[])+[(u[i]-1,a[i])]
  p2[u[i]-1]=p2.get(u[i]-1,[])+[(v[i]-1,b[i])]
  p2[v[i]-1]=p2.get(v[i]-1,[])+[(u[i]-1,b[i])]

d = dijkstra(p,s-1,n)
d2 = dijkstra(p2,t-1,n)

ans = [0]*n
ans[0] = d[n-1]+d2[n-1]
for i in range(1,n):
  ans[i] = min(ans[i-1],d[n-1-i]+d2[n-1-i])

for i in ans[::-1]:
  print(10**15-i)