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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <iostream>
#include <queue>
#include <vector>
using namespace std;

#define INF 1e+9
#define MAX_V 100005

struct edge {
    int to;
    int cost;
};

// <最短距離, 頂点の番号>
using P = pair<int, int>;

int V;
vector<edge> G[MAX_V];
vector<edge> rG[MAX_V];
int d[MAX_V];
int rd[MAX_V];

void dijkstra(int s) {
    priority_queue<P, vector<P>, greater<P> > que;
    fill(d, d+V, INF);
    d[s] = 0;
    que.push(P(0, s));

    while (!que.empty()) {
        P p = que.top();
        que.pop();
        int v = p.second;
        if (d[v] < p.first) continue;

        for (int i=0; i<G[v].size(); ++i) {
            edge e = G[v][i];
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                que.push(P(d[e.to], e.to));
            }
        }
    }
}

void rdijkstra(int s) {
    priority_queue<P, vector<P>, greater<P> > que;
    fill(rd, rd+V, INF);
    rd[s] = 0;
    que.push(P(0, s));

    while (!que.empty()) {
        P p = que.top();
        que.pop();
        int v = p.second;
        if (rd[v] < p.first) continue;

        for (int i=0; i<rG[v].size(); ++i) {
            edge e = rG[v][i];
            if (rd[e.to] > rd[v] + e.cost) {
                rd[e.to] = rd[v] + e.cost;
                que.push(P(rd[e.to], e.to));
            }
        }
    }
}

int main() {
    cin >> V;
    int E,s,t;
    cin >> E >> s >> t;
    for (int i=0; i<E; ++i) {
        int a, b, c, d;
        cin >> a >> b >> c >> d;
        edge e1 = {b-1, c};
      	edge e2 = {a-1, c};
      	edge re1 = {b-1, d};
      	edge re2 = {a-1, d};
      	rG[b-1].push_back(re2);
      	rG[a-1].push_back(re1);
        G[a-1].push_back(e1);
      	G[b-1].push_back(e2);
    }
    dijkstra(s-1);
  	rdijkstra(t-1);
  	priority_queue<int, vector<int>, greater<int>> q;
  	vector<long long int> ans(V);
  	long long int pt = 1e15;
	for(int i=0;i<V;i++){
    	q.push(d[V-i-1]+rd[V-i-1]);
      	ans[V-i-1] = pt - q.top();
    }
  	for(auto x: ans) cout << x << endl;
  	return 0;
}