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++14 (GCC 5.4.1) (judge verdict: Accepted)

//INCLUDE
//------------------------------------------
#include <bits/stdc++.h>
//DEFINE
//------------------------------------------
#define ll long long
#define ld long double
#define ALLv(a)  (a).begin(),(a).end()
#define ALL(a,n)  (a),(a)+n
#define vi vector<long long>
#define vd vector<long double>
#define vs vector<string>

//CONST
//------------------------------------------
#define INF 1010000000000000017LL
#define MOD 1000000007LL
#define EPS 1e-12
#define PI 3.14159265358979323846
//REPEAT
//------------------------------------------
#define   FOR(i,m,n) for(ll (i)=(m);   (i)<(n);  (i)++)
#define   REP(i,n)   for(ll (i)=0;     (i)<(n);  (i)++)
#define  REPS(i,x)   for(ll (i)=1;     (i)<=(x); (i)++)
#define  RREP(i,x)   for(ll (i)=(x)-1; (i)>=0;   (i)--)
#define RREPS(i,x)   for(ll (i)=(x);   (i)> 0;   (i)--)
#define  WREP(i,in,j,jn) REP(i,in)REP(j,jn)
//-----------------------------------------
#define ifYN(x) cout<<(x?"Yes":"No")<<"\n" 
#define dcml(n) fixed<<setprecision(n)

 
ll GCD(ll a,ll b) { return b ? GCD(b,a%b) : a; } 
ll LCM(ll a,ll b) { return a/GCD(a,b)*b; } 
ll Factorial(ll n){ ll m=1;while(n>=1)m*=n--;return m; } 
ll modPow(ll x,ll n,ll mod=MOD){ 
    return (!n?1:!n%2?modPow(x*x,n/2,mod)%mod:x%mod*modPow(x,n-1,mod)%mod); 
} 
ll GetDigit(ll num){ 
    ll digit=0; 
    while(num!=0){num/=10;digit++;} 
    return digit; 
} 


using namespace std;
//-------------------------------------------
typedef pair<ll,ll> P;
ll n;
struct Edge{
  ll to,dist;
};
ll dy[200000];
ll ds[200000];
vector<Edge> Gy[200000];
vector<Edge> Gs[200000];
void dijkstraY(ll u){
  REP(i,n)dy[i]=INF;
  priority_queue<P,vector<P>,greater<P>> que;
  que.push({0,u});
  while(!que.empty()){
    auto r=que.top();que.pop();
    if(r.first>=dy[r.second])continue;
    dy[r.second]=r.first;
    
    for(Edge& e: Gy[r.second]){
      ll dist=dy[r.second]+e.dist;
      if(dist>=dy[e.to])continue;
      que.push({r.first+e.dist,e.to});
    }
  }
}
void dijkstraS(ll u){
  REP(i,n)ds[i]=INF;
  priority_queue<P,vector<P>,greater<P>> que;
  que.push({0,u});
  while(!que.empty()){
    auto r=que.top();que.pop();
    if(r.first>=ds[r.second])continue;
    ds[r.second]=r.first;
    
    for(Edge& e: Gs[r.second]){
      ll dist=ds[r.second]+e.dist;
      if(dist>=ds[e.to])continue;
      que.push({r.first+e.dist,e.to});
    }
  }
}
int main(void){
  cin.tie(0);
  ios::sync_with_stdio(false);
  
  ll m,s,t;cin>>n>>m>>s>>t;//n都市,m電車,都市sからtに移動
  s--,t--;
  ll u0,v0,a0,b0;
  REP(i,m){
    cin>>u0>>v0>>a0>>b0;//a円bスヌーク
    u0--,v0--;
    Gy[u0].push_back({v0,a0});
    Gy[v0].push_back({u0,a0});
    Gs[u0].push_back({v0,b0});
    Gs[v0].push_back({u0,b0});
  }
  dijkstraY(s);
  dijkstraS(t);
  
  ll ans[n+1];ans[n]=INF;
  REP(i,n){
    ans[n-i-1]=min(ans[n-i],dy[n-i-1]+ds[n-i-1]);
  }
  REP(i,n){
    cout<<(1000000000000000LL-ans[i])<<"\n";
  }
  return 0;
}