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: Wrong Answer)

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

typedef struct{
    long long q;
    long long i;
}queue_t;

int main(void){
    long long n,p,s,t,count;
    scanf("%lld%lld%lld%lld",&n,&p,&s,&t);

    long long mm=0,head[3*p+1],next[3*p+1],to[3*p+1],leny[3*p+1],lens[3*p+1];
    for (int i=0; i<=n; i++){
        head[i]=-1;
    }
    for (long long i=0; i<p; i++){
        long long u,v,a,b;
        scanf("%lld%lld%lld%lld",&u,&v,&a,&b);
        next[mm]=head[u];
        head[u]=mm;
        to[mm]=v;
        leny[mm]=a;
        lens[mm]=b;
        mm++;
        next[mm]=head[v];
        head[v]=mm;
        to[mm]=u;
        leny[mm]=a;
        lens[mm]=b;        
        mm++;
    }
    
    count=1;
    queue_t q[n*50],tmpq;
    long long d[n+1];
    for (long long i=0; i<=n; i++){
        d[i]=1e17;
        q[i].q=d[i];
        q[i].i=i;
        count++;
    }
    d[s]=0;
    q[count].q=d[s];
    q[count].i=s;
    for (long long i=count; i>0; i=i/2){
        if (i/2!=0 && q[i].q<=q[i/2].q){
            tmpq=q[i];
            q[i]=q[i/2];
            q[i/2]=tmpq;
        } else {
            break;
        }
    }
    for (long long i=0; i<=n; i++){
        long long now=q[1].i;
        if (now==t){
            break;
        }
        q[1]=q[count];
        long long bf=1;
        while(1){
            if (bf*2>count){
                break;
            }
            if (q[bf*2].q<=q[bf*2+1].q){
                if (q[bf].q>q[bf*2].q){
                    tmpq=q[bf];
                    q[bf]=q[bf*2];
                    q[bf*2]=tmpq;
                    bf=bf*2;
                } else {
                    break;
                }
            } else {
                if (q[bf].q>q[bf*2+1].q){
                    tmpq=q[bf];
                    q[bf]=q[bf*2+1];
                    q[bf*2+1]=tmpq;
                    bf=bf*2+1;
                } else {
                    break;
                }
            }
        }
        count--;
        long long tmp;
        for (long long e=head[now]; e!=-1; e=next[e]){
            tmp=d[now]+leny[e];
            if (d[to[e]]>tmp){
                d[to[e]]=tmp;
                count++;
                q[count].q=tmp;
                q[count].i=to[e];
                for (long long af=count; af>0; af=af/2){
                    if (af/2!=0 && q[af].q<=q[af/2].q){
                        tmpq=q[af];
                        q[af]=q[af/2];
                        q[af/2]=tmpq;
                    } else {
                        break;
                    }
                }
            }
        }
    }
    
    count=1;
    queue_t q2[n*50];
    long long f[n+1];
    for (long long i=0; i<=n; i++){
        f[i]=1e17;
        q2[i].q=f[i];
        q2[i].i=i;
        count++;
    }
    f[t]=0;
    q2[count].q=f[t];
    q2[count].i=t;
    for (long long i=count; i>0; i=i/2){
        if (i/2!=0 && q2[i].q<=q2[i/2].q){
            tmpq=q2[i];
            q2[i]=q2[i/2];
            q2[i/2]=tmpq;
        } else {
            break;
        }
    }
    for (long long i=0; i<=n; i++){
        int now=q2[1].i;
        if (now==s){
            break;
        }
        q2[1]=q2[count];
        long long bf=1;
        while(1){
            if (bf*2>=count){
                break;
            }
            if (q2[bf*2].q<=q2[bf*2+1].q){
                if (q2[bf].q>q2[bf*2].q){
                    tmpq=q2[bf];
                    q2[bf]=q2[bf*2];
                    q2[bf*2]=tmpq;
                    bf=bf*2;
                } else {
                    break;
                }
            } else {
                if (q2[bf].q>q2[bf*2+1].q){
                    tmpq=q2[bf];
                    q2[bf]=q2[bf*2+1];
                    q2[bf*2+1]=tmpq;
                    bf=bf*2+1;
                } else {
                    break;
                }
            }
        }
        count--;
        long long tmp;
        for (long long e=head[now]; e!=-1; e=next[e]){
            tmp=f[now]+lens[e];
            if (f[to[e]]>tmp){
                f[to[e]]=tmp;
                count++;
                q2[count].q=tmp;
                q2[count].i=to[e];
                for (long long af=count; af>0; af=af/2){
                    if (af/2!=0 && q2[af].q<=q2[af/2].q){
                        tmpq=q2[af];
                        q2[af]=q2[af/2];
                        q2[af/2]=tmpq;
                    } else {
                        break;
                    }
                }
            }
        }
    }
    
    long long ans[n+1];
    ans[0]=0;
    for (int i=1; i<=n; i++){
        ans[i]=d[i]+f[i];
    }
    for (int i=n; i>1; i--){
        if(ans[i]<ans[i-1]){
            ans[i-1]=ans[i];
        }
    }
    for (int i=1; i<=n; i++){
        printf("%lld\n",1000000000000000-ans[i]);
    }
    
	return 0 ;
}