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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
long long R=1,C=1,H[2000010],N[2000010];
//評価関数（いまはMIN）
int hyouka(int a,int b){
  if(C<b)return 1;
  if(C<a||b==0)return 0;
  return N[H[a]]<N[H[b]]?1:0;
}
//挿入関数
void hin(long long a){
  int i=C++;
  for(N[H[0]=R]=a;hyouka(0,i/2);i/=2)H[i]=H[i/2];
  H[i]=R++;
}
//取り出す関数
int hout(){
  int rt=H[1],i,j=2,k=H[--C];
  for(i=1;hyouka(i,C);i=j)H[i]=H[j=i*2+1-hyouka(i*2,i*2+1)];
  H[j/2]=k;
  return rt;
}
long long CO[100010];
void dijk(int v,int e,int *fr,int *to,int *co,int mi){
  int i,id[100010],ta[100010],nt[100010],f[100010]={0};
  for(i=0;i<=v;i++)ta[i]=-1;
  for(i=0;i<=v;i++)CO[i]=1000000000;
  for(i=CO[mi]=0;i<e;i++){
    nt[i]=ta[fr[i]];
    ta[fr[i]]=i;
  }
  while(f[mi]-1){
    f[mi]=1;
    for(i=ta[mi];i+1;i=nt[i]){
      if(CO[to[i]]>CO[mi]+co[i])hin(CO[id[R]=to[i]]=CO[mi]+co[i]);
    }
    while(f[mi]&&C-1)mi=id[hout()];
  }
}
long long MIN(long long a,long long b){return a<b?a:b;}
int main(){
  int n,m,a[200010],b[200010],c[200010],d[200010],s,t,i;
  long long e[100010],f[100010];
  scanf("%d %d %d %d",&n,&m,&s,&t);
  for(i=0;i<m;i++)scanf("%d %d %d %d",&a[i],&b[i],&c[i],&d[i]);
  for(i=0;i<m;i++){
    a[i+m]=b[i];
    b[i+m]=a[i];
    c[i+m]=c[i];
    d[i+m]=d[i];
  }
  dijk(n,m*2,a,b,c,s);
  for(i=R=C=1;i<=n;i++)e[i]=CO[i];
  dijk(n,m*2,a,b,d,t);
  f[n+1]=1e18;
  for(i=n;i;i--)f[i]=MIN(f[i+1],e[i]+CO[i]);
  for(i=0;i<n;i++)printf("%lld\n",(long long)1e15-f[i+1]);
  return 0;
}