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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
int n,m,s,t;
template<typename T> using min_priority_queue=priority_queue<T,vector<T>,greater<T>>;
//template<typename T> using min_priority_queue=priority_queue<T,vector<T>,greater<T>>;
int vis[100000];
void dijk(int s,vector<vector<pair<int,int>>> &E,vector<ll> &D){
    for(int i=0;i<n;i++){
        D[i]=1LL<<60;
        vis[i]=0;
    }
    min_priority_queue<pair<ll,int>> que;
    D[s]=0;
    que.push({0,s});
    while(!que.empty()){
        auto p=que.top();
        que.pop();
        ll cnt=p.first;
        int lo=p.second;
        if(vis[lo])continue;
        vis[lo]=1;
        for(auto x:E[lo]){
            ll cnt2=cnt+x.second;
            int to=x.first;
            if(D[to]>cnt2){
                D[to]=cnt2;
                que.push({D[to],to});
            }
        }
    }
}
ll dp[100000];
int main(void){
    // Your code here!
    cin>>n>>m>>s>>t;
    s--,t--;
    vector<vector<pair<int,int>>> EA(n);
    vector<vector<pair<int,int>>> EB(n);
    vector<ll> DA(n),DB(n);
    for(int i=0;i<m;i++){
        int u,v,a,b;
        cin>>u>>v>>a>>b;
        u--,v--;
        EA[u].push_back({v,a});
        EA[v].push_back({u,a});
        EB[u].push_back({v,b});
        EB[v].push_back({u,b});
    }
    dijk(s,EA,DA);
    dijk(t,EB,DB);
    for(int i=0;i<n;i++){
        dp[i]=DA[i]+DB[i];
    }
    //cout<<DA[0]<<DB[0]<<endl;
    for(int i=n-2;i>=0;i--){
        dp[i]=min(dp[i],dp[i+1]);
    }
    for(int i=0;i<n;i++){
        cout<<1000000000000000LL-dp[i]<<endl;
    }
}