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>
#include<stdlib.h>
#include<math.h>

typedef long long int int64;

#define MAX(a,b) ((a)>(b)?(a):(b))
#define MIN(a,b) ((a)<(b)?(a):(b))
#define ABS(a) ((a)>(0)?(a):-(a))

typedef struct dnode{
  int v;
  int64 val;
} node;

int cmp(node *a,node *b){
  return a->val<=b->val?1:0;
}

void swap(node *a,node *b){
  node t=*a;
  *a=*b;
  *b=t;
  return;
}

void push(node *h,int v,int64 val){
  h[++h[0].v].v=v;
  h[h[0].v].val=val;
  int k=h[0].v;
  while(k>1){
    if(cmp(h+k/2,h)) break;
    swap(h+k/2,h+k);
    k/=2;
  }
  return;
}

int pop(node *h){
  int res=h[1].v;
  h[1]=h[h[0].v--];
  int k=1;
  while(2*k+1<=h[0].v){
    if(cmp(h+k,h+2*k) && cmp(h+k,h+2*k+1)) break;
    int next=cmp(h+2*k,h+2*k+1)?2*k:2*k+1;
    swap(h+k,h+next);
    k=next;
  }
  if(2*k<=h[0].v && !cmp(h+k,h+2*k)) swap(h+k,h+2*k);
  return res;
}

void dijkstra(int from,int64 *array,int *start,int *vertex,int *cost,int *next,int n){
  int *used=(int *)malloc(sizeof(int)*(n+1));
  int i;
  for(i=0;i<=n;i++){
    array[i]=10000000000000000LL;
    used[i]=0;
  }
  node *h=(node *)malloc(sizeof(node)*100001);
  h[0].v=0;
  array[from]=0;
  push(h,from,0);
  while(h[0].v>0){
    int v=pop(h);
    if(used[v]) continue;
    used[v]=1;
    int p=start[v];
    while(p>0){
      int u=vertex[p];
      if(array[v]+cost[p]<array[u]){
	array[u]=array[v]+cost[p];
	push(h,u,array[u]);
      }
      p=next[p];
    }
  }
  free(used);
  free(h);
  return;
}

#define ADD(i,u,v,a,b) do{vertex[i]=v;costA[i]=a;costB[i]=b;next[i]=start[u];start[u]=i;}while(0)

void run(void){
  int n,m,s,t;
  scanf("%d%d%d%d",&n,&m,&s,&t);
  int i;
  int *start=(int *)malloc(sizeof(int)*(n+1));
  int *vertex=(int *)malloc(sizeof(int)*(2*m+1));
  int *costA=(int *)malloc(sizeof(int)*(2*m+1));
  int *costB=(int *)malloc(sizeof(int)*(2*m+1));
  int *next=(int *)malloc(sizeof(int)*(2*m+1));
  next[0]=0;
  for(i=0;i<=n;i++) start[i]=0;
  for(i=1;i<=m;i++){
    int u,v,a,b;
    scanf("%d%d%d%d",&u,&v,&a,&b);
    ADD(2*i-1,u,v,a,b);
    ADD(2*i,v,u,a,b);
  }
  int64 *sTo=(int64 *)malloc(sizeof(int64)*(n+1));
  dijkstra(s,sTo,start,vertex,costA,next,n);
  int64 *tTo=(int64 *)malloc(sizeof(int64)*(n+1));
  dijkstra(t,tTo,start,vertex,costB,next,n);
  int64 *min=(int64 *)malloc(sizeof(int64)*(n+1));
  min[n]=sTo[n]+tTo[n];
  for(i=n-1;i>=1;i--){
    min[i]=MIN(min[i+1],sTo[i]+tTo[i]);
  }
  const int64 mm=1000000000000000LL;
  for(i=1;i<=n;i++){
    printf("%lld\n",mm-min[i]);
  }
}

int main(void){
  run();
  return 0;
}