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 (Clang 3.8.0) (judge verdict: Wrong Answer)

#include <algorithm>
#include <iomanip>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <string>
#include <utility>
#include <vector>

using Path = std::pair<long long, int>;
using Graph = std::vector<std::vector<Path>>;
using List = std::vector<long long>;

List find_min(const Graph &g, int s) {
  int n = g.size();
  std::priority_queue<Path, std::vector<Path>, std::greater<Path>> heap;
  List dist(n, 1e18);
  dist[s] = 0;
  heap.push({0, s});
  while (not heap.empty()) {
    long long t = heap.top().first;
    int pos = heap.top().second;
    heap.pop();
    if (t != dist[pos]) continue;
    for (Path p : g[pos]) {
      long long t2 = t + p.first;
      long long v = p.second;
      if (t2 < dist[v]) {
        dist[v] = t2;
        heap.push({t2, v});
      }
    }
  }
  return dist;
}

int main() {
  int n, m, s, t;
  std::cin >> n >> m >> s >> t;
  Graph ga(n), gb(n);
  for (int mi = 0; mi < m; ++mi) {
    int u, v, a, b;
    std::cin >> u >> v >> a >> b;
    ga[u - 1].push_back({a, v - 1});
    ga[v - 1].push_back({a, u - 1});
    gb[u - 1].push_back({b, v - 1});
    gb[v - 1].push_back({b, u - 1});
  }
  List da = find_min(ga, s - 1);
  List db = find_min(gb, t - 1);
  List ans(n);
  ans[n - 1] = da[n - 1] + db[n - 1];
  for (int i = n - 2; i >= 0; --i) ans[i] = std::min(ans[i + 1], da[i] + db[i]);
  for (long long x : ans) std::cout << x << std::endl;
  for (long long x : ans) std::cout << 1000000000000000 - x << std::endl;
  return 0;
}