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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

/*
考え方としては、
i年に移動する際には、必ず一度はi以降の都市に立ち寄らないといけない
その時に立ち寄る都市をYとすると、
sからYまでにかかった円
Yからtまでにかかったすぬーく
の合計金額が最も小さくなるような方法をとればいい。
したがって、
sから各都市までにかかる円の最小値
tから各都市までにかかるすぬーくの最小値
を求めて、i以降の都市でその合計値が最小になるようなものを選べばよい。
*/
struct edge{
    int64_t next_pos, yen, snook;
};
/*
struct yen_edge{
    int64_t  yen, next_pos;
};
struct snook_edge{
    int64_t  snook, next_pos;
};
bool operator<(const yen_edge &left, const yen_edge &right) {
    return left.yen < right.yen;
}

bool operator<(const snook_edge &left, const snook_edge &right) {
    return left.snook < right.snook;
}
*/

int main(){
    int64_t n, m, s, t;
    cin>> n >> m >> s >> t;

    vector<vector<edge>> G(n+1);
    for(int64_t i=0;i<m;i++){
        int64_t u,v,a,b;
        cin>>u>>v>>a>>b;
        edge s1 = {v,a,b};
        edge s2 = {u,a,b};
        G.at(u).push_back(s1);
        G.at(v).push_back(s2);
    }

    //ダイクストラ法でスタート地点sから各都市までにかかる最小の円を求める
    //ダイクストラ法にはheapを用いる
    priority_queue<pair<int64_t,int64_t>,vector<pair<int64_t,int64_t>>,greater<pair<int64_t,int64_t>>> heap1;
    vector<int64_t> check_list1(n+1,0); 
    vector<int64_t> yen_min_map(n+1,INT_MAX); 

    yen_min_map.at(s) = 0;
    check_list1.at(s) = 1;
    int64_t current_pos = s;
    //cout<<current_pos<<endl;
    while(true){
        vector<edge> edge_list = G.at(current_pos);
        int64_t current_yen = yen_min_map.at(current_pos);

        //cout<<"edge_size:"<<edge_list.size()<<endl;
        for(int64_t i=0;i<edge_list.size();i++){
            edge a = edge_list.at(i);
            //cout<<"i:"<<i<<",nextpos:"<<a.next_pos<<"yen:"<<a.yen<<endl;

            if(check_list1.at(a.next_pos)==1){
//                continue;
            }
            pair<int64_t, int64_t> b = {a.yen+current_yen,a.next_pos};
            heap1.push(b);
        }

            if(heap1.size()==0){
                goto OUT;
            }
        int64_t next_node, cost_yen;
        while(true){
            pair<int64_t,int64_t> selected_edge = heap1.top();
            heap1.pop();
            tie(cost_yen,next_node) = selected_edge;
            //cout<<next_node<<","<<cost_yen<<endl;
            if(check_list1.at(next_node)!=1){
                break;
            }
            if(heap1.size()==0){
                goto OUT;
            }
        }
        //cout<<"next_node:"<<next_node<<endl;
        yen_min_map[next_node] =cost_yen;
                 
        check_list1.at(next_node) = 1;

        current_pos = next_node;
    }
OUT:

    priority_queue<pair<int64_t,int64_t>,vector<pair<int64_t,int64_t>>,greater<pair<int64_t,int64_t>>> heap2;
    vector<int64_t> check_list2(n+1,0); 
    vector<int64_t> snook_min_map(n+1,INT_MAX); 

    snook_min_map.at(t) = 0;
    check_list2.at(t) = 1;
    current_pos = t;
    //cout<<current_pos<<endl;
    while(true){
        vector<edge> edge_list = G.at(current_pos);
        int64_t current_snook = snook_min_map.at(current_pos);

        //cout<<"edge_size:"<<edge_list.size()<<endl;
        for(int64_t i=0;i<edge_list.size();i++){
            //cout<<"i:"<<i<<endl;
            edge a = edge_list.at(i);
            if(check_list2.at(a.next_pos)==1){
                continue;
            }
            pair<int64_t, int64_t> b = {a.snook+current_snook,a.next_pos};
            heap2.push(b);
        }

            if(heap2.size()==0){
                goto OUT2;
            }
        int64_t next_node, cost_snook;
        while(true){
            pair<int64_t,int64_t> selected_edge = heap2.top();
            heap2.pop();
            tie(cost_snook,next_node) = selected_edge;
            //cout<<next_node<<","<<cost_snook<<endl;
            if(check_list2.at(next_node)!=1){
                break;
            }
            if(heap2.size()==0){
                goto OUT2;
            }
        }
        //cout<<"next_node:"<<next_node<<endl;
        snook_min_map[next_node] =cost_snook;
                 
        check_list2.at(next_node) = 1;

        current_pos = next_node;
    }


OUT2:


for(int64_t i=0;i<yen_min_map.size();i++){
    //cout<<"yen:"<<i<<" "<<yen_min_map.at(i)<<endl;
}

for(int64_t i=0;i<snook_min_map.size();i++){
    //cout<<"snook:"<<i<<" "<<snook_min_map.at(i)<<endl;
}


vector<int64_t> total_cost(n+1,INT_MAX);
for(int64_t i=1;i<n+1;i++){
    total_cost.at(i) = yen_min_map.at(i) + snook_min_map.at(i);
}

int64_t first_money = pow(10,15);
for(int64_t i=0;i<n;i++){
    int64_t min = *std::min_element(total_cost.begin()+i+1, total_cost.end());
    cout<<first_money-min<<endl;
}







return 0;



}