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;

#define int long long

struct Dijkstra {
	struct E {
		int t; // to
		int c; // cost
		E(int t_, int c_) : t(t_), c(c_) {}
//		E(void) : t(0), c(0) {}
		inline bool operator<(const E a) const {
			return c < a.c;
		}
		inline bool operator>(const E a) const {
			return a < (*this);
		}
	};
	int n; // num of vertex
	vector<vector<E>> es; // edges list
	Dijkstra(int sz) : n(sz), es(sz) {}
    void add_edge(int s, int t, int cost) {
        es[s].push_back(E(t, cost));
    }
	// s : start, inf : infinity
    auto calc(int s, int inf) {
        vector<int> res(es.size(), inf);
		// res[i] : cost of position i
        priority_queue<E, vector<E>, greater<E>> q;
        q.emplace(s, 0);
        while (!q.empty()) {
            auto a = q.top();
            q.pop();
			// a.t : now position
			// a.c : now position's cost
            if (a.c >= res[a.t]) continue;
            res[a.t] = a.c;
            for (auto& p : es[a.t]) {
                if (a.c + p.c < res[p.t]) q.emplace(p.t, a.c + p.c);
            }
        }
        return res;
    }
};

signed main()
{
	int n, m, S, T;
	cin >> n >> m >> S >> T;
	S--, T--;
	Dijkstra EN(n), SN(n);
	for (int i = 0; i < m; i++) {
		int v, u, x, y;
		cin >> v >> u >> x >> y;
		v--, u--;
		EN.add_edge(u, v, x);
		EN.add_edge(v, u, x);
		SN.add_edge(u, v, y);
		SN.add_edge(v, u, y);
	}
	auto en = EN.calc(S, LLONG_MAX);
	auto sn = SN.calc(T, LLONG_MAX);
	vector<int> res(n);
	for (int i = 0; i < n; i++) res[i] = en[i] + sn[i];
	for (int i = n - 2; i >= 0; i--) res[i] = min(res[i], res[i + 1]);
	for (int i = 0; i < n; i++) cout << (int)1e15 - res[i] << endl;
}
/*
	verified on 2018/5/13
	https://atcoder.jp/contests/soundhound2018-summer-qual/tasks/soundhound2018_summer_qual_d
*/