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.

Examples

Input

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1


Output

999999999999998
999999999999989
999999999999979
999999999999897


Input

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


Output

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
#ダイクストラ
import heapq
n,m,s,t = map(int,input().split())
UVAB = [list(map(int,input().split())) for _ in [0]*m]

Ea = [{} for _ in [0]*n]
Eb = [{} for _ in [0]*n]
for u,v,a,b in UVAB:
    Ea[u-1][v-1] = a
    Ea[v-1][u-1] = a
    Eb[u-1][v-1] = b
    Eb[v-1][u-1] = b

#import heapq
def Dijkstra(N,E,start):
    """
    N: ノード数
    E: 辺(dictのlist)
    start:開始ノード番号(0-indexed)
    """
    d = [-1]*N
    q = [(0,start-1)]
    while q:
        ci,i = heapq.heappop(q)
        if d[i]!= -1:continue
        d[i] = ci
        for j,cj in E[i].items():
            if d[j]!=-1:continue
            heapq.heappush(q,(ci+cj,j))
    return d

da = Dijkstra(n,Ea,s)
db = Dijkstra(n,Eb,t)

C = [a+b for a,b in zip(da,db)]

ans = [C[-1]]
for c in C[-2::-1]:
    ans.append(min(ans[-1],c))

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