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: Runtime Error)

#include "bits/stdc++.h"
 
typedef long long ll;
#define int ll
#define fi first
#define se second
#define SORT(a) sort(a.begin(),a.end())
#define rep(i,n) for(int i = 0;i < (n) ; i++) 
#define REP(i,n) for(int i = 0;i < (n) ; i++) 
#define MP(a,b) make_pair(a,b)
#define pb(a) push_back(a)
#define INF LLONG_MAX/2
#define all(x) (x).begin(),(x).end()
#define debug(x) cerr<<#x<<": "<<x<<endl
#define debug_vec(v) cerr<<#v<<":";rep(i,v.size())cerr<<" "<<v[i];cerr<<endl
using namespace std;
int MOD = 1000000007;
int n,m,s,t;
#define MAX_V 10010
struct edge{int to,cost;};
typedef pair<int,int> pii;
int V;
vector<edge> G[2][MAX_V];
int d[2][MAX_V];
void dijkstra(int s,int x){
    priority_queue<pii,vector<pii>,greater<pii>> que;
    rep(i,V)d[x][i] = INF;
    //fill(d[x],d[x]+V,INF);
    d[x][s] = 0;
    que.push(pii(0,s));
    while(!que.empty()){
        pii p = que.top();que.pop();
        int v = p.se;
        if(d[x][v] < p.fi) continue;
        rep(i,G[x][v].size()){
            edge e = G[x][v][i];
            if(d[x][e.to] > d[x][v]+e.cost){
                d[x][e.to] = d[x][v] + e.cost;
                que.push(pii(d[x][e.to],e.to));
            }
        }
    }
}
vector<pair<int,int>> cst;

signed main(){
    cin >> n >> m >> s >> t;
    s--;
    t--;
    V = n;
    rep(i,m){//m は edge数
        int u,v,a,b;
        cin >> u >> v >> a >> b;
      	u--;
      	v--;
      	edge e1 = {u,a};
      	edge e2 = {v,a};
        G[0][v].pb(e1);
        G[0][u].pb(e2);
        edge e3 = {u,b};
      	edge e4 = {v,b};
        G[1][v].pb(e3);
        G[1][u].pb(e4);
    }
    dijkstra(s,0);//yen
    dijkstra(t,1);//snuke

    rep(i,n){
        cst.pb(MP(d[0][i]+d[1][i],i));
    }

    SORT(cst);
    int num = 0;
    rep(i,n){
        while(cst[num].se < i){
            num++;
        }
        cout << 1000000000000000-cst[num].fi << endl;
    }
    
    return 0; 
}

// g++ -std=c++14 code1.cpp
// rm -r -f test;oj dl https://arc064.contest.atcoder.jp/tasks/arc064_c
// rm -r -f test;oj dl http://soundhound2018-summer-qual.contest.atcoder.jp/tasks/soundhound2018_summer_qual_d