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 collections import defaultdict
import heapq

def dijkstra(G, s):
    dist = [float('inf')] * len(G.keys())
    dist[s] = 0
    q = [(dist[v], v) for v in G.keys()]
    heapq.heapify(q)
    while True:
        try:
            _, u = heapq.heappop(q)
        except IndexError:
            break
        for v, d in G[u]:
            tmp = dist[u] + d
            if dist[v] > tmp:
                dist[v] = tmp
                heapq.heappush(q, (tmp, v))
    return dist

n, m, s, t = map(int, input().split())
G1 = defaultdict(list)
G2 = defaultdict(list)

for i in range(m):
    u, v, a, b = map(int, input().split())
    u -= 1
    v -= 1
    G1[u].append((v, a))
    G1[v].append((u, a))
    G2[u].append((v, b))
    G2[v].append((u, b))

dist1 = dijkstra(G1, s-1)
dist2 = dijkstra(G2, t-1)

mv = 0
ans = []
for i in range(n-1, -1, -1):
    mv = max(mv, 10**15 - dist1[i] - dist2[i])
    ans.append(mv)

ans.reverse()
for a in ans:
    print(a)