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 (Clang 3.8.0) (judge verdict: Time Limit Exceeded)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <stdbool.h>
#include <math.h>
#include <limits.h>

int main(void){
    long long n,m,s,t;
    scanf("%lld%lld%lld%lld",&n,&m,&s,&t);
    s--;
    t--;
    long long u[m],v[m],a[m],b[m];
    for (int i=0; i<m; i++){
        scanf("%lld%lld%lld%lld",&u[i],&v[i],&a[i],&b[i]);
        u[i]--;
        v[i]--;
    }
    long long y[m],k[m];
    bool visited[n];
    for (int i=0; i<n; i++){
        y[i]=LLONG_MAX;
        visited[i]=false;
    }    
    y[s]=0;
    for (int i=0; i<n; i++){
        long long tmp=LLONG_MAX,next;
        for (int l=0; l<n; l++){
            if (tmp>y[l] && visited[l]==false){
                tmp=y[l];
                next=l;
            }
        }
        visited[next]=true;
        for (int l=0; l<m; l++){
            if (u[l]==next){
                if (y[v[l]]>y[next]+a[l]){
                    y[v[l]]=y[next]+a[l];
                }
            }
            if (v[l]==next){
                if (y[u[l]]>y[next]+a[l]){
                    y[u[l]]=y[next]+a[l];
                }
            }
        }
    }
    for (int i=0; i<n; i++){
        k[i]=LLONG_MAX;
        visited[i]=false;
    }    
    k[t]=0;
    for (int i=0; i<n; i++){
        long long tmp=LLONG_MAX,next;
        for (int l=0; l<n; l++){
            if (tmp>k[l] && visited[l]==false){
                tmp=k[l];
                next=l;
            }
        }
        visited[next]=true;
        for (int l=0; l<m; l++){
            if (u[l]==next){
                if (k[v[l]]>k[next]+b[l]){
                    k[v[l]]=k[next]+b[l];
                }
            }
            if (v[l]==next){
                if (k[u[l]]>k[next]+b[l]){
                    k[u[l]]=k[next]+b[l];
                }
            }
        }
    } 
    long long ans[n];
    for (int i=0; i<n; i++){
        ans[i]=y[i]+k[i];
    }
    for (int i=n-1; i>0; i--){
        if (ans[i-1]>ans[i]){
            ans[i-1]=ans[i];
        }
    }
    for (int i=0; i<n; i++){
        printf("%lld\n",1000000000000000-ans[i]);
    }
	return 0 ;
}