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;
using ll = long long;
const ll INF = 1e18;

vector<ll> dijkstra(int N,int start,int dest,const vector<vector<pair<ll,ll>>>& Hen){
    vector<ll> dist(N, INF), prever(N, -1), seen(N);
    dist[start] = 0;
    priority_queue<pair<ll,ll>, vector<pair<ll,ll>>, greater<pair<ll,ll>>> que;
    que.emplace(0, start);

    while(que.size()) {
        ll ver = que.top().second, cost = que.top().first; que.pop();
        if(seen[ver])continue;
        seen[ver] = true;
        for(auto i : Hen[ver]) {
            if(dist[i.first] > cost + i.second) {
                dist[i.first] = cost + i.second;
                prever[i.first] = ver;
                que.emplace(dist[i.first], i.first);
            }
        }
    }
    return dist;
}


int main()
{
    int n, m, s, t; cin >> n >> m >> s >> t; s--, t--;
    vector<vector<pair<ll,ll>>> en(n), sn(n);
    for(int i = 0; i < m; i++) {
        int u, v, a, b;
        cin >> u >> v >> a >> b; u--, v--;
        en[u].emplace_back(v, a); en[v].emplace_back(u, a);
        sn[u].emplace_back(v, b); sn[v].emplace_back(u, b);
    }
    vector<ll> sta = dijkstra(n, s, t, en);
    vector<ll> end = dijkstra(n, t, s, sn);

    ll nax = 0, def = 1e15;
    vector<ll> cost(n);
    for(int i = n - 1; i >= 0; i--) {
        nax = max(nax, def - (sta[i] + end[i]));
        cost[i] = nax;
    }
    for(int i = 0; i < n; i++)cout << cost[i] << endl;
}