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;
#define rep(i,a) for(int i=0;i<(a);i++)
const ll MOD=1000000007;

struct edge {int to; ll cost;};
typedef pair<int, int> P; // firstは最短距離, secondは頂点の番号

const int INF=INT_MAX/2;
const int MAX_V=101010;
int V,E;
vector<vector<edge> > G1(MAX_V),G2(MAX_V);
ll d[2][MAX_V];

void dijkstra1(int s){
  int x=0;
  priority_queue<P, vector<P>, greater<P> > que;
  rep(j,MAX_V) d[x][j]=INF;
  d[x][s]=0;
  que.push(P(0, s));
  while(!que.empty()){
    P p = que.top(); que.pop();
    int v = p.second; // 頂点の番号
    if (d[x][v] < p.first) continue;
    for(int i=0;i<G1[v].size();i++){
      edge e = G1[v][i];
      if (d[x][e.to] > d[x][v] + e.cost){
        d[x][e.to] = d[x][v] + e.cost;
        que.push(P(d[x][e.to], e.to));
      }
    }
  }
}

void dijkstra2(int s){
  int x=1;
  priority_queue<P, vector<P>, greater<P> > que;
  rep(j,MAX_V) d[x][j]=INF;
  d[x][s]=0;
  que.push(P(0, s));
  while(!que.empty()){
    P p = que.top(); que.pop();
    int v = p.second; // 頂点の番号
    if (d[x][v] < p.first) continue;
    for(int i=0;i<G2[v].size();i++){
      edge e = G2[v][i];
      if (d[x][e.to] > d[x][v] + e.cost){
        d[x][e.to] = d[x][v] + e.cost;
        que.push(P(d[x][e.to], e.to));
      }
    }
  }
}

int main(){
  int s,t;
  cin>>V>>E>>s>>t; s--,t--;
  rep(i,E){
    int u,v,a,b; cin>>u>>v>>a>>b; u--,v--;
    G1[u].push_back({v,a});
    G2[u].push_back({v,b});
    G1[v].push_back({u,a});
    G2[v].push_back({u,b});
  }
  dijkstra1(s);
  dijkstra2(t);
  ll hav=1e15;
  priority_queue<pair<ll,int> > pq;
  rep(i,V){
    ll tmp=hav;
    tmp-=d[0][i];
    tmp-=d[1][i];
    pq.push(make_pair(tmp,i));
  }
  rep(i,V){
    pair<ll,int> p=pq.top();
    while(p.second<i){
      pq.pop();
      p=pq.top();
    }
    cout<<p.first<<endl;
  }
  return 0;
}