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: Accepted)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<assert.h>

typedef struct binaryHeap{
  void *h;
  int heapSize;
  int maxSize;
  size_t size;
  int (*cmp)(const void *,const void *);
} heap;

heap* newHeap(size_t size,int (*cmpF)(const void *,const void *)){
  const int maxSize=1;
  heap *res=(heap *)malloc(sizeof(heap));
  res->h=malloc(size*(maxSize+1));
  res->heapSize=0;
  res->maxSize=maxSize;
  res->size=size;
  res->cmp=cmpF;
  return res;
}

int isEmpty(heap *h){
  return h->heapSize==0;
}

void freeHeap(heap *h){
  free(h->h);
  free(h);
  return;
}

inline void heapFunc_swap(void * restrict a,void * restrict b,void * restrict tmp,size_t size){
  memcpy(tmp,a,size);
  memcpy(a,b,size);
  memcpy(b,tmp,size);
  return;
}

void push(heap *q,void *p){
  if(q->heapSize==q->maxSize){
    q->h=realloc(q->h,q->size*(2*q->maxSize+1));
    assert(q->h!=NULL);
    q->maxSize*=2;
  }
  q->heapSize+=1;
  void *h=q->h;
  int k=q->heapSize;
  size_t size=q->size;
  int (*cmp)(const void *,const void *)=q->cmp;
  memcpy(h+k*size,p,size);
  while(k>1){
    if(cmp(h+(k/2)*size,h+k*size)) return;
    heapFunc_swap(h+(k/2)*size,h+k*size,h,size);
    k/=2;
  }
  return;
}

void pop(heap *q,void *res){
  assert(q->heapSize>0);
  void *h=q->h;
  size_t size=q->size;
  memcpy(res,h+size,size);
  memcpy(h+size,h+size*q->heapSize,size);
  q->heapSize-=1;
  int (*cmp)(const void *,const void *)=q->cmp;
  int n=q->heapSize;
  int k=1;
  while(2*k+1<=n){
    int next=cmp(h+2*k*size,h+(2*k+1)*size)?2*k:2*k+1;
    if(cmp(h+k*size,h+next*size)) return;
    heapFunc_swap(h+k*size,h+next*size,h,size);
    k=next;
  }
  if(2*k<=n && cmp(h+2*k*size,h+k*size)) heapFunc_swap(h+k*size,h+2*k*size,h,size);
  return;
}

typedef long long int int64;

typedef struct costNode{
  int index;
  int64 cost;
} node;

int cmp(const void *a,const void *b){
  return ((node *)a)->cost<=((node *)b)->cost;
}

void dijkstra(int from,int *vertex,int *cost,int *next,int *start,int64 *dp,int n){
  const int64 inf=1000000000000000LL;
  heap *h=newHeap(sizeof(node),cmp);
  for(int i=0;i<n;i++) dp[i]=inf;
  dp[from]=0;
  push(h,&((node){from,0}));
  while(!isEmpty(h)){
    node t;
    pop(h,&t);
    int v=t.index;
    if(dp[v]<t.cost) continue;
    for(int p=start[v];p!=-1;p=next[p]){
      int u=vertex[p];
      int64 m=dp[v]+cost[p];
      if(m>=dp[u]) continue;
      dp[u]=m;
      push(h,&((node){u,m}));
    }
  }
  freeHeap(h);
  return;
}

#define ADD(u,v,a,b,index) do{vertex[index]=u;costA[index]=a;costB[index]=b;next[index]=start[v];start[v]=index;}while(0)
void run(void){
  int n,m,s,t;
  scanf("%d%d%d%d",&n,&m,&s,&t);
  s--;t--;
  int *vertex=(int *)malloc(sizeof(int)*2*m);
  int *next=(int *)malloc(sizeof(int)*2*m);
  int *costA=(int *)malloc(sizeof(int)*2*m);
  int *costB=(int *)malloc(sizeof(int)*2*m);
  int *start=(int *)malloc(sizeof(int)*n);
  int i;
  for(i=0;i<n;i++) start[i]=-1;
  for(i=0;i<m;i++){
    int u,v,a,b;
    scanf("%d%d%d%d",&u,&v,&a,&b);
    u--;v--;
    ADD(u,v,a,b,2*i);
    ADD(v,u,a,b,2*i+1);
  }
  int64 *dpA=(int64 *)malloc(sizeof(int64)*n);
  int64 *dpB=(int64 *)malloc(sizeof(int64)*n);
  dijkstra(s,vertex,costA,next,start,dpA,n);
  dijkstra(t,vertex,costB,next,start,dpB,n);
  int64 *ans=(int64 *)malloc(sizeof(int64)*n);
  const int64 ini=1000000000000000LL;
  ans[n-1]=ini-(dpA[n-1]+dpB[n-1]);
  for(i=n-2;i>=0;i--){
    int64 m=ini-(dpA[i]+dpB[i]);
    ans[i]=m>=ans[i+1]?m:ans[i+1];
  }
  for(i=0;i<n;i++) printf("%lld\n",ans[i]);
}

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