p03305 SoundHound Inc. Programming Contest 2018 -Masters Tournament- - Saving Snuuk
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
```python
import math, string, itertools, fractions, collections, re, array, bisect, sys, random, time, copy, functools
from heapq import heappush, heappop, heappushpop, heapify, heapreplace

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')

```