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: Time Limit Exceeded)

#pragma region header
#include <bits/stdc++.h>
#define int ll
#define all(a) begin(a), end(a)
#define rall(a) rbegin(a), rend(a)
#define mp make_pair
#define rep1(i, n) for (decltype(+n) i = 0; i < (n); i++)
#define rrep1(i, n) for (auto i = n - 1; i > static_cast<decltype(i)>(-1); i--)
#define rep2(i, a, b) for (auto i = (a); i < (b); i++)
#define rrep2(i, a, b) for (auto i = b - 1; i >= a; i--)
#define GET_MACRO(_1, _2, _3, NAME, ...) NAME
#define rep(...) GET_MACRO(__VA_ARGS__, rep2, rep1) (__VA_ARGS__)
#define rrep(...) GET_MACRO(__VA_ARGS__, rrep2, rrep1) (__VA_ARGS__)
#define each(i, a) for (auto &&i : (a))
using namespace std;

using ll = long long;

const int INF = 1e18;

template <class T>
bool amin(T &a, const T &b) {
  if (a > b) {
    a = b;
    return true;
  }
  return false;
}

#pragma endregion header

template<typename T>
struct edge {
  int to;
  T cost;
  edge(int to, T cost) : to(to), cost(cost) {}
};
template<typename T>
vector<T> dijkstra(vector<vector<edge<T>>> &g, int s, const T &INF) {
  vector<T> dist(g.size(), INF);
  priority_queue<pair<T, int>> pq;
  dist[s] = 0;
  pq.emplace(dist[s], s);
  while(!pq.empty()) {
    T cost;
    int idx;
    tie(cost, idx) = pq.top();
    pq.pop();
    if(dist[idx] < cost) continue;
    for(auto &e : g[idx]) {
      auto next_cost = cost + e.cost;
      if(dist[e.to] <= next_cost) continue;
      dist[e.to] = next_cost;
      pq.emplace(dist[e.to], e.to);
    }
  }
  return dist;
}

const int MAX = 1e15;

#pragma region main
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);

  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--;t--;
  vector<vector<edge<int>>> g(n), rg(n);
  rep(_, m) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--;v--;
    g[u].emplace_back(v, a);
    g[v].emplace_back(u, a);
    rg[u].emplace_back(v, b);
    rg[v].emplace_back(u, b);
  }
  vector<int> dij = dijkstra(g, s, INF), rdij = dijkstra(rg, t, INF);
  rep(i, n) dij[i] += rdij[i];
  rrep(i, n - 1) amin(dij[i], dij[i + 1]);
  rep(i, n) cout << MAX - dij[i] << '\n';
}
#pragma endregion main