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;

struct edge { long long to, cost;};
typedef pair<long long, long long> P_dij;

vector<long long> dijkstra(long long s, long long n, vector<edge> G[]){
    vector<long long> d = {};
    long long INF = 10000000000000000;
    d.resize(n);
    for(int i=0; i<n; i++){
        d[i] = INF;
    }
    priority_queue<P_dij, vector<P_dij>, greater<P_dij>> que;
    d[s] = 0;
    que.push(P_dij(0, s));
    while(!que.empty()){
        P_dij p = que.top();
        que.pop();
        long long 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_dij(d[e.to], e.to));
            }
        }
    }
    return d;
}

long long i, j, k;
vector<edge> edge_yen[200001];
vector<edge> edge_snu[200001];

int main(){
    long long n, m, s, t;
    cin >> n >> m >> s >> t;
    for(i=0; i<m; i++){
        long long u, v, a, b;
        cin >> u >> v >> a >> b;
        edge e1;
        e1.to = v, e1.cost = a;
        edge_yen[u].push_back(e1);
        edge e2;
        e2.to = u, e2.cost = a;
        edge_yen[v].push_back(e2);
        edge e3;
        e3.to = v, e3.cost = b;
        edge_snu[u].push_back(e3);
        edge e4;
        e4.to = u, e4.cost = b;
        edge_snu[v].push_back(e4);
    }

    vector<long long> dist_yen = dijkstra(s, n+1, edge_yen);
    vector<long long> dist_snu = dijkstra(t, n+1, edge_snu);

    long long ans[200001];
    long long Max = 0;
    long long FIRST = 1e15;
    for(i=n; i>=1; i--){
        long long val = FIRST - (dist_yen[i] + dist_snu[i]);
        Max = max(val, Max);
        ans[i] = Max;
    }
    for(i=1; i<=n; i++){
        cout << ans[i] << endl;
    }
    return 0;
}