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 heapq

n, m, s, t = map(int, input().split())
adjYen = [[] for v in range(n)]
adjSnuuk = [[] for v in range(n)]

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

# 単一始点最短経路を求める（ダイクストラ法）
def Dijkstra(adjList, vSt):
    numV = len(adjList)

    numDone = 0
    pq = []

    cost = [float('inf')] * numV
    cost[vSt] = 0
    heapq.heappush(pq, (0, vSt))

    while pq:
        c, vNow = heapq.heappop(pq)
        if c > cost[vNow]: continue

        numDone += 1
        if numDone == numV: break

        for v2, wt in adjList[vNow]:
            c2 = cost[vNow] + wt
            if c2 < cost[v2]:
                cost[v2] = c2
                heapq.heappush(pq, (c2, v2))

    return cost


costYen = Dijkstra(adjYen, s-1)
costSnuuk = Dijkstra(adjSnuuk, t-1)

cost = [(y + s, v) for v, (y, s) in enumerate(zip(costYen, costSnuuk))]
cost.sort()

iC = 0
for i in range(n):
    while cost[iC][1] < i:
        iC += 1
    print(10 ** 15 - cost[iC][0])
