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++ (GCC 9.2.1) (judge verdict: Accepted)

#include <iostream>
#include <queue>
#include <utility>
#include <vector>

using namespace std;

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;

    vector<vector<pair<int, long long>>> ga(n, vector<pair<int, long long>>(0));
    vector<vector<pair<int, long long>>> gb(n, vector<pair<int, long long>>(0));
    for (int i = 0; i < m; i++) {
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        ga[u - 1].push_back(make_pair(v - 1, a));
        ga[v - 1].push_back(make_pair(u - 1, a));
        gb[u - 1].push_back(make_pair(v - 1, b));
        gb[v - 1].push_back(make_pair(u - 1, b));
    }

    long long inf = 1LL << 61;

    vector<long long> ds(n, inf);
    ds[s - 1] = 0;
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> ques;
    ques.push(make_pair(0, s - 1));
    while (!ques.empty()) {
        int q = ques.top().second;
        ques.pop();

        for (int i = 0; i < static_cast<int>(ga[q].size()); i++) {
            if (ds[ga[q][i].first] > ds[q] + ga[q][i].second) {
                ds[ga[q][i].first] = ds[q] + ga[q][i].second;
                ques.push(make_pair(ds[ga[q][i].first], ga[q][i].first));
            }
        }
    }

    vector<long long> dt(n, inf);
    dt[t - 1] = 0;
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> quet;
    quet.push(make_pair(0, t - 1));
    while (!quet.empty()) {
        int q = quet.top().second;
        quet.pop();

        for (int i = 0; i < static_cast<int>(gb[q].size()); i++) {
            if (dt[gb[q][i].first] > dt[q] + gb[q][i].second) {
                dt[gb[q][i].first] = dt[q] + gb[q][i].second;
                quet.push(make_pair(dt[gb[q][i].first], gb[q][i].first));
            }
        }
    }

    vector<long long> res(n, ds[n - 1] + dt[n - 1]);
    for (int i = n - 2; i >= 0; i--) {
        res[i] = min(res[i + 1], ds[i] + dt[i]);
    }

    for (int i = 0; i < n; i++) {
        cout << 1000000000000000LL - res[i] << endl;
    }
}