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 <bits/stdc++.h>
using namespace std;
typedef long long ll;

struct edge{int to,cost;};
typedef pair<ll,int> P;

vector<vector<edge> > G(100000);
vector<ll> d(100000);

vector<vector<edge> > G2(100000);
vector<ll> d2(100000);

vector<ll> d3(100000);

int n,m,s,t;

ll INF=100000000000;

void dijkstra(int s,vector<vector<edge> >& G,vector<ll>& d){
  priority_queue<P,vector<P>,greater<P> > que;
  fill(d.begin(),d.begin()+n,INF);
  d[s]=0;
  que.push(P(0,s));
  while(!que.empty()){
    P p=que.top();que.pop();
    int v = p.second;
    if(d[v]<p.first) continue;
    for(int i=0;i<G[v].size();i++){
      edge e = G[v][i];
      if(d[e.to]>d[v]+e.cost){
        d[e.to]=d[v]+e.cost;
        que.push(P(d[e.to],e.to));
      }
    }
  }
}

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 e = {v,a};
    G[u].push_back(e);
    e.cost=b;
    G2[u].push_back(e);
    e.to=u;
    e.cost=a;
    G[v].push_back(e);
    e.cost=b;
    G2[v].push_back(e);
  }
  dijkstra(s,G,d);
  dijkstra(t,G2,d2);

  for(int i=0;i<n;i++){
    d3.at(i)=d.at(i)+d2.at(i);
  }
  vector<ll> d4(100000);
  d4.at(0)=d3.at(n-1);
  for(int i=1;i<n-1;i++){
    ll mini = d4.at(i-1);
    if(d3.at(n-1-i)<mini){
      mini=d3.at(n-1-i);
    }
    d4.at(i)=mini;
  }
  for(int i=1;i<n;i++){
    cout << 1000000000000000-d4.at(n-i-1) << endl;
  }
}