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<iomanip>
#include<string>
#include<vector>
#include<queue>
#include<set>
#include<map>
#include<bitset>
#include<time.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
#define INF (1<<30)
#define INFLL (1LL << 60)
#define MOD 1000000007
#define rep(i, n) for(int i=0; i<(n); ++i)
template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; }return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; }return false; }

struct Edge {
	int to, cost;
	Edge(int t, int c) { to = t; cost = c; }
};
using Pi = pair<int, int>;
using Graph = vector<vector<Edge>>;
vector<int> dijkstra(const Graph& graph, int start)
{
	priority_queue<Pi, vector<Pi>, greater<Pi>> pq;	//コスト、頂点
	pq.push(make_pair(0, start));
	vector<int> cost(graph.size(), INF);
	cost[start] = 0;
	while (!pq.empty()) {
		auto top = pq.top(); pq.pop();
		int c = top.first;
		int v = top.second;
		if (c > cost[v]) continue;
		cost[v] = c;
		for (Edge e : graph[v]) {
			int nc = c + e.cost;
			if (nc >= cost[e.to]) continue;
			pq.push(make_pair(nc, e.to));
		}
	}
	return cost;
}

int main()
{
	const ll have = 1000000000000000;
	int N, M, S, T; cin >> N >> M >> S >> T; --S; --T;
	Graph Yen(N);
	Graph Snk(N);
	rep(i, M) {
		int u, v, a, b; cin >> u >> v >> a >> b;--u;--v;
		Yen[u].emplace_back(Edge(v, a));
		Yen[v].emplace_back(Edge(u, a));
		Snk[u].emplace_back(Edge(v, b));
		Snk[v].emplace_back(Edge(u, b));
	}
	auto yen = dijkstra(Yen, S);
	auto snk = dijkstra(Snk, T);
	vector<int> used(N, INF);
	int minuse = INF;
	for (int year = N - 1; year >= 0; --year) {
		chmin(minuse, yen[year] + snk[year]);
		used[year] = minuse;
	}
	rep(i, N) cout << have - used[i] << endl;
	return 0;
}