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
import math, string, itertools, fractions, collections, re, array, bisect, sys, random, time, copy, functools
from heapq import heappush, heappop, heappushpop, heapify, heapreplace
#input = sys.stdin.readline

N, M, S, T = [int(_) for _ in input().split()]
UVAB = [[int(_) for _ in input().split()] for _ in range(M)]
G1 = collections.defaultdict(lambda: collections.defaultdict(int))
G2 = collections.defaultdict(lambda: collections.defaultdict(int))

for u, v, a, b in UVAB:
    u -= 1
    v -= 1
    G1[u][v] = G1[v][u] = a
    G2[u][v] = G2[v][u] = b


def dijkstra(V, G, start):
    INF = float('inf')
    D = [INF for _ in range(V)]
    D[start] = 0
    H = [(0, start)]
    while H:
        du, u = heappop(H)
        if D[u] < du:
            continue
        for v, d in G[u].items():
            if D[v] > du + d:
                D[v] = du + d
                heappush(H, (D[v], v))
    return D


D1 = dijkstra(N, G1, S - 1)
D2 = dijkstra(N, G2, T - 1)

D = [10**15 - sum(_) for _ in zip(D1, D2)]
for i in range(N - 1, 0, -1):
    D[i - 1] = max(D[i - 1], D[i])
print(*D, sep='\n')
