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: Wrong Answer)

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

using int64 = long long;
using State = pair<int64, int>; // <cost, city>

int main() {
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  --s;
  --t;
  // [JPY(0)/SNK(1)][from] = [(to,cost)]
  vector<vector<vector<pair<int, int>>>> costs(2, vector<vector<pair<int,int>>>(n));
  for (int i = 0; i < m; ++i) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    --u;
    --v;
    costs[0][u].push_back({v, a});
    costs[0][v].push_back({u, a});
    costs[1][u].push_back({v, b});
    costs[1][v].push_back({u, b});
  }

  vector<vector<int64>> to_city(2, vector<int64>(n, -1));
  set<State> q;
  q.insert({0LL, s});
  to_city[0][s] = 0;
  // dijkstra only for JPY
  while (!q.empty()) {
    auto front = q.begin();
    int64 cost = front->first;
    int city = front->second;
    q.erase(front);
    for (auto p : costs[0][city]) {
      int to = p.first;
      int64 sum = cost + p.second;
      if (to_city[0][to] >= 0 && to_city[0][to] < sum)
        continue;
      to_city[0][to] = sum;
      q.insert({sum, to});
    }
  }

  // move to SNK
  vector<int64> ret(n, 1e+15);
  for (int i = n - 1; i >= 0; --i) {
    q.insert({to_city[0][i], i});
    to_city[1][i] = to_city[0][i];
    while (!q.empty()) {
      auto front = q.begin();
      int64 cost = front->first;
      int city = front->second;
      q.erase(front);
      for (auto p : costs[1][city]) {
        int to = p.first;
        int64 sum = cost + p.second;
        if (to_city[1][to] >= 0 && to_city[1][to] < sum)
          continue;
        to_city[1][to] = sum;
        q.insert({sum, to});
      }
    }
    ret[i] = ret[i] - to_city[1][t];
  }
  for (int64 r : ret)
    cout << r << "\n";

  return 0;
}