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++ (Clang 10.0.0) (judge verdict: Wrong Answer)

#include<iostream>
#include<vector>
#include<string>
#include<algorithm>
#include<map>
#include<set>
#include<queue>
#define INFLL (1LL << 60)
#define INF (1 << 30)
using namespace std;
typedef long long ll;

namespace Dijkstra
{
	struct Edge {
		int to, weight;
		Edge(int t, int w) {
			to = t; weight = w;
		}
	};
	using Graph = vector<vector<Edge>>;
	vector<ll> dijkstra(int s, const Graph& graph) {
		const int N = graph.size();
		priority_queue<pair<int, int>> pq;	//コスト、頂点
		pq.push(make_pair(0, s));
		vector<ll> cost(N, INFLL);
		cost[s] = 0;
		while (!pq.empty()) {
			auto t = pq.top(); pq.pop();
			int c = t.first;
			int v = t.second;
			for (auto e : graph[v]) {
				int nc = cost[v] + e.weight;
				if (nc > cost[e.to]) continue;	//探索済みよりコストが大きいとき、探索省略
				pq.push(make_pair(nc, e.to));
				cost[e.to] = nc;
			}
		}
		return cost;
	}
	void exec()
	{
		const ll have = 1000000000000000;
		int N, M, S, T; cin >> N >> M >> S >> T; --S; --T;
		Graph yen(N), snk(N);
		for (int i = 0; i < M; ++i) {
			int u, v, a, b; cin >> u >> v >> a >> b; --u, --v;
			yen[u].push_back(Edge(v, a));
			yen[v].push_back(Edge(u, a));
			snk[u].push_back(Edge(v, b));
			snk[v].push_back(Edge(u, b));
		}
		auto c1 = dijkstra(S, yen);
		auto c2 = dijkstra(T, snk);
		vector<ll> anss(N, INFLL);
		for (int year = 0; year < N; ++year) {
			for (int i = year; i < N; ++i) {
				ll r = c1[i] + c2[i];
				anss[year] = min(anss[year], r);
			}
			cout << have - anss[year] << endl;
		}
	}
}

int main()
{
	Dijkstra::exec();
	return 0;
}