Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

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

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include "bits/stdc++.h"
using namespace std;

typedef pair<long long, long long> P;
long long INF = 1145141919810364364;

int main() {
	int N, M, S, T;
	static long long YenDis[100000], SnuukDis[100000], ANS[100000];
	static vector<P> Yen[100000], Snuuk[100000];
	priority_queue<P, vector<P>, greater<P> > PQYen, PQSnuuk;
	cin >> N >> M >> S >> T;
	S--, T--;
	for (int i = 0; i < N; i++) {
		YenDis[i] = INF;
		SnuukDis[i] = INF;
	}
	YenDis[S] = 0;
	SnuukDis[T] = 0;
	for (int i = 0; i < M; i++) {
		long long U, V, A, B;
		cin >> U >> V >> A >> B;
		U--, V--;
		Yen[U].push_back(P(A, V));
		Yen[V].push_back(P(A, U));
		Snuuk[U].push_back(P(B, V));
		Snuuk[V].push_back(P(B, U));
	}
	for (P p : Yen[S]) PQYen.push(p), YenDis[p.second] = p.first;
	for (P p : Snuuk[T]) PQSnuuk.push(p), SnuukDis[p.second] = p.first;
	while (PQYen.size()) {
		P p = PQYen.top();
		PQYen.pop();
		for (P q : Yen[p.second]) {
			if (YenDis[q.second] > YenDis[p.second] + q.first) {
				YenDis[q.second] = YenDis[p.second] + q.first;
				PQYen.push(q);
			}
		}
	}
	while (PQSnuuk.size()) {
		P p = PQSnuuk.top();
		PQSnuuk.pop();
		for (P q : Snuuk[p.second]) {
			if (SnuukDis[q.second] > SnuukDis[p.second] + q.first) {
				SnuukDis[q.second] = SnuukDis[p.second] + q.first;
				PQSnuuk.push(q);
			}
		}
	}
	ANS[0] = YenDis[N - 1] + SnuukDis[N - 1];
	for (int i = 1; i < N; i++) {
		ANS[i] = min(ANS[i - 1], YenDis[N - 1 - i] + SnuukDis[N - i - 1]);
	}
	for (int i = N - 1; i >= 0; i--) {
		cout << 1000000000000000 - ANS[i] << endl;
	}
}