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 <iostream>
#include <string>
#include <vector>
#include <climits>
#include <algorithm>
#include <set>
#include <unordered_map>
#include <queue>
#include <iomanip>

using namespace std;
int n, m, s, t;
vector<long> dist1(100005);
vector<long> dist2(100005);
vector<long> u(100005);
vector<long> v(100005);
vector<long> a(100005);
vector<long> b(100005);
vector<vector<pair<long, long>>> graph(100005);

void dijkstra1(long s) {
  priority_queue<pair<long,long>, vector<pair<long, long>>, greater<>> queue;
  queue.push(make_pair(0, s));
  while (!queue.empty()) {
    pair<long, long> p = queue.top();
    queue.pop();
    long start = p.second;
    long d = p.first;
    if (d > dist1.at(start)) {
      continue;
    }
    for (auto e : graph.at(start)) {
      long end = e.first;
      long train = e.second;
      long cost = a.at(train);
      if (dist1.at(end) > d + cost) {
        dist1.at(end) = d + cost;
        // cout << "yen: pos: " << end << " cost: " << dist1.at(end) << endl;
        queue.push(make_pair(dist1.at(end), end));
      }
    }
  }
}

void dijkstra2(long t) {
  priority_queue<pair<long,long>, vector<pair<long, long>>, greater<>> queue;
  queue.push(make_pair(0, t));
  while (!queue.empty()) {
    pair<long, long> p = queue.top();
    queue.pop();
    long start = p.second;
    long d = p.first;
    if (d > dist2.at(start)) {
      continue;
    }
    for (auto e : graph.at(start)) {
      long end = e.first;
      long train = e.second;
      long cost = b.at(train);
      if (dist2.at(end) > d + cost) {
        dist2.at(end) = d + cost;
        // cout << "snuke: pos: " << end << " cost: " << dist2.at(end) << endl;
        queue.push(make_pair(dist2.at(end), end));
      }
    }
  }
}

int main() {
  cin >> n >> m >> s >> t;
  for (int i=1; i<=m; i++) {
    cin >> u.at(i) >> v.at(i) >> a.at(i) >> b.at(i);
  }

  for (int i=1; i<=m; i++) {
    graph.at(u.at(i)).push_back(make_pair(v.at(i), i));
    graph.at(v.at(i)).push_back(make_pair(u.at(i), i));
  }

  for (int i=1; i<=n; i++) {
    dist1.at(i) = LONG_MAX;
    dist2.at(i) = LONG_MAX;
  }
  dist1.at(s) = 0;
  dist2.at(t) = 0;

  dijkstra1(s);
  dijkstra2(t);

  vector<long> dvec(n+1);
  for (int i=1; i<=n; i++) {
    dvec.at(i) = dist1.at(i) + dist2.at(i);
  }
  for (int i=n-1; i>=1; i--) {
    dvec.at(i) = min(dvec.at(i), dvec.at(i+1));
  }
  long init = 1;
  for (int i=1; i<=15; i++) {
    init *= 10;
  }
  for (int i=1; i<=n; i++) {
    cout << init-dvec.at(i) << endl;
  }
}