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;
typedef long long ll;
#define rep(i, n) for(ll i = 0, i##_len = (n); i < i##_len; i++)
#define reps(i, s, n) for(ll i = (s), i##_len = (n); i < i##_len; i++)
#define rrep(i, n) for(ll i = (n) - 1; i >= 0; i--)
#define rreps(i, e, n) for(ll i = (n) - 1; i >= (e); i--)
#define all(x) (x).begin(), (x).end()
#define sz(x) ((ll)(x).size())
#define len(x) ((ll)(x).length())

template<class T>
struct Dijkstra {
    const T inf = numeric_limits<T>::max() / 2 - 1;
    vector<T> dist;
    
    Dijkstra() {}
    Dijkstra(int n) {
        resize(n);
    }
    
    void resize(int n) {
        g.resize(n);
        dist.resize(n);
        used.resize(n);
    }
    
    void add_edge(int a, int b, T c) {
        g[a].emplace_back(c, b);
    }
    
    void get_distance(int a) {
        priority_queue<pair<T, int>, vector<pair<T, int>>, greater<pair<T, int>>> pq;
        fill(dist.begin(), dist.end(), inf);
        fill(used.begin(), used.end(), false);
        pq.push(make_pair(0, a));
        while(!pq.empty()) {
            T d;
            int t;
            tie(d, t) = pq.top();
            pq.pop();
            if (used[t]) continue;
            used[t] = true;
            dist[t] = d;
            for (auto e : g[t]) {
                if (dist[e.second] <= (d + e.first)) continue;
                pq.push(make_pair(d + e.first, e.second));
            }
        }
    }
    
private:
    vector<vector<pair<int, T>>> g;
    vector<bool> used;
};

struct S {
    ll u, v, a, b;
};

int main() {
    cin.tie(0);
    ios::sync_with_stdio(false);
    // std::ifstream in("input.txt");
    // std::cin.rdbuf(in.rdbuf());
    const ll inf = LONG_LONG_MAX / 2 - 1;
    ll n, m, s, t;
    cin >> n >> m >> s >> t;
    s--; t--;
    vector<S> uvab(m);
    rep(i, m) {
        cin >> uvab[i].u >> uvab[i].v >> uvab[i].a >> uvab[i].b;
        uvab[i].u--; uvab[i].v--;
    }
    Dijkstra<ll> yen(n);
    rep(i, m) {
        yen.add_edge(uvab[i].u, uvab[i].v, uvab[i].a);
        yen.add_edge(uvab[i].v, uvab[i].u, uvab[i].a);
    }
    yen.get_distance(s);
    Dijkstra<ll> snuke(n);
    rep(i, m) {
        snuke.add_edge(uvab[i].u, uvab[i].v, uvab[i].b);
        snuke.add_edge(uvab[i].v, uvab[i].u, uvab[i].b);
    }
    snuke.get_distance(t);
    priority_queue<pair<ll, ll>, vector<ll>, greater<ll>> pq;
    vector<ll> ans(n);
    rrep(i, n) {
        pq.push(yen.dist[i] + snuke.dist[i]);
        ans[i] = 1e15 - pq.top();
    }
    rep(i, n) {
        cout << ans[i] << endl;
    }
    return 0;
}