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<cstdio>
#include<queue>
#include<algorithm>
#include<vector>
using namespace std;


int n, m, s, t;
vector<pair<int, long long> > edge[2][100000];
long long dist[2][100000], dist_res[100000];
long long ans[100000];

void dijkstra(int kind, int start){
  priority_queue<pair<long long, int> > Q;

  for(int i = 0; i < 100000; i++){
    dist[kind][i] = 10000000000000000;
  }
  dist[kind][start] = 0;
  Q.push(make_pair(-0, start));
  while(!Q.empty()){
    pair<long long, int> q = Q.top(); Q.pop();
    int now = q.second;
    long long d = -(q.first);
    if(dist[kind][now] <= d){
      continue;
    }
    dist[kind][now] = d;
    for(int i = 0; i < edge[kind][now].size(); i++){
      if(edge[kind][now][i].second + d < dist[kind][edge[kind][now][i].first]){
	dist[kind][edge[kind][now][i].first] = edge[kind][now][i].second + d;
	Q.push(make_pair(-dist[kind][edge[kind][now][i].first], edge[kind][now][i].first));
      }
    }
  }
}


int main(){
  cin >> n >> m >> s >> t;
  s--; t--;
  for(int i = 0; i < m; i++){
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--;
    v--;
    edge[0][u].push_back(make_pair(v, a));
    edge[0][v].push_back(make_pair(u, a));
    edge[1][u].push_back(make_pair(v, b));
    edge[1][v].push_back(make_pair(u, b));
  }

  dijkstra(0, s);
  dijkstra(1, t);
  for(int i = 0; i < n; i++){
    dist_res[i] = dist[0][i] + dist[1][i];
  }
  ans[n-1] = dist_res[n-1];
  for(int i = n-1; i > 0; i--){
    ans[i-1] = min(ans[i], dist_res[i-1]);
  }
  for(int i = 0; i < n; i++){
    printf("%lld\n", 1000000000000000 - ans[i]);
  }
}