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
from sys import stdin, stdout
from heapq import heappush, heappop
from itertools import repeat
def main():
    n, m, s, t = map(int, stdin.readline().split())
    dat = map(int, stdin.read().split(), repeat(10, 4 * m))
    la = [None] * (n + 1)
    xt = [None] * (4 * m)
    j = 0
    for i in xrange(m):
        x, y = dat[j], dat[j+1]
        xt[j] = la[y]
        la[y] = j
        j += 1
        xt[j] = la[x]
        la[x] = j
        j += 3
    dd = [[10 ** 15] * (n + 1), [10 ** 15] * (n + 1)]
    for i in xrange(2):
        d = dd[i]
        x = t if i else s
        d[x] = 0
        q = []
        heappush(q, (0, x))
        while q:
            co, x = heappop(q)
            y = la[x]
            if d[x] != co:
                continue
            while y is not None:
                z = dat[y]
                yy = y / 4 * 4 + 2
                c = dat[yy+i]
                if d[z] > co + c:
                    d[z] = co + c
                    heappush(q, (d[z], z))
                y = xt[y]
    ans = [0] * n
    z = 10 ** 15
    ans[n-1] = z - dd[0][n] - dd[1][n]
    for i in xrange(n - 2, -1, -1):
        ans[i] = x = z - dd[0][i+1] - dd[1][i+1]
        if ans[i+1] > x:
            ans[i] = ans[i+1]
    stdout.write(''.join('%d\n' % x for x in ans))
main()
