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

vector<int64_t> dsy(vector<int64_t>dis,vector<vector<pair<int64_t,int64_t>>> g,int64_t start,int64_t goal){
    int n=dis.size();
    priority_queue<pair<int64_t,int64_t>,vector<pair<int64_t,int64_t>>,greater<pair<int64_t,int64_t>>> Q;
    dis[start]=0;
    Q.push(make_pair(0,start));

    while(!Q.empty()){
        int pos=Q.top().second;Q.pop();
        for(int i=0;i<g.at(pos).size();i++){
            int to=g.at(pos).at(i).first;
            int cost=g.at(pos).at(i).second;
            if(dis[to]==-1||dis[to]>dis[pos]+cost){
                dis[to]=dis[pos]+cost;
                Q.push(make_pair(dis[to],to));
            }
        }
    }
    return dis;
}

int64_t dss(vector<int64_t>dis,vector<vector<pair<int64_t,int64_t>>> g,int64_t start,int64_t goal){
    int n=dis.size();
    priority_queue<pair<int64_t,int64_t>,vector<pair<int64_t,int64_t>>,greater<pair<int64_t,int64_t>>> Q;
    dis[start]=0;
    Q.push(make_pair(0,start));

    while(!Q.empty()){
        int pos=Q.top().second;Q.pop();
        for(int i=0;i<g.at(pos).size();i++){
            int to=g.at(pos).at(i).first;
            int cost=g.at(pos).at(i).second;
            if(dis[to]==-1||dis[to]>dis[pos]+cost){
                dis[to]=dis[pos]+cost;
                Q.push(make_pair(dis[to],to));
            }
        }
    }
    return dis[goal];
}

int main(){
    int64_t n,m,s,t; cin>>n>>m>>s>>t;
    int64_t money=pow(10,15);

    vector<vector<pair<int64_t,int64_t>>>gy(n),gs(n);
    vector<int64_t> dis(n,-1);
    vector<int64_t> dis2(n,-1);

    for(int i=0;i<m;i++){
        int u,v,a,b;
        cin>>u>>v>>a>>b;
        gy.at(u-1).push_back(make_pair(v-1,a));
        gy.at(v-1).push_back(make_pair(u-1,a));
        gs.at(u-1).push_back(make_pair(v-1,b));
        gs.at(v-1).push_back(make_pair(u-1,b));
    }

    dis=dsy(dis,gy,s-1,t-1);//sから各国への円の合計
    dis2=dsy(dis2,gs,t-1,s-1);//tから各国へのすぬけの合計
    vector<int64_t> dis_min(n);
    dis_min.at(n-1)=dis[n-1]+dis2[n-1];

    for(int i=n-2;i>=0;i--){
        dis_min[i]=min(dis[i]+dis2[i],dis_min[i+1]);
    }

    for(int i=0;i<n;i++){
        cout<<money-dis_min[i]<<endl;
    }
    /*
    for(int i=0;i<n;i++){//各年でのループ
        int64_t result=pow(10,16);
        for(int j=0;j<n;j++){
            result=min(result,dis[i]+dis2[i]);
        }
        
        for(int j=i;j<n;j++){//どの国で両替するか
            int64_t disj=dis[j];
            int64_t dssj=dss(dis2,gs,j,t-1);
            if(disj==-1||dssj==-1) continue;
            int64_t sum=disj+dssj;
            //cout << disj<<" "<<dssj<<" "<<sum << endl;
            result=min(result,sum);
        }
        cout << money-result << endl;
    }*/
}