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: Wrong Answer)

#include "bits/stdc++.h"
using namespace std;
#define MOD 1000000007
#define INF 9223372036854775807
#define FOR(i,a,b) for(long long i=(a);i<(b);i++)
#define RFOR(i,a,b) for(long long i = (b-1);i>=a;i--)
#define REP(i,n)  FOR(i,0,n)
#define RREP(i,n) RFOR(i,0,n)
#define ITR(itr,mp) for(auto itr = (mp).begin(); itr != (mp).end(); ++itr)
#define RITR(itr,mp) for(auto itr = (mp).rbegin(); itr != (mp).rend(); ++itr)
#define dump(x)  cout << #x << " = " << (x) << endl;
#define debug(x) cout << #x << " = " << (x) << " (L" << __LINE__ << ")" << " " << __FILE__ << endl;
typedef long long ll;
typedef pair<ll,ll> P;
typedef vector<P> Graph;

ll n,m,s,t;
const int NMAX = 100001; 
struct node{
  ll op,a,b;
};

ll dista[NMAX];
ll distb[NMAX];
vector<node> nodes[NMAX];
bool visited[NMAX];

void dfsa(ll id){
  REP(i,nodes[id].size()){
    dista[nodes[id][i].op] = min(dista[nodes[id][i].op],dista[id]+nodes[id][i].a);
    if(visited[nodes[id][i].op]) continue;
    visited[nodes[id][i].op] = true;
    dfsa(nodes[id][i].op);
  }
}

void dfsb(ll id){
  REP(i,nodes[id].size()){
    distb[nodes[id][i].op] = min(distb[nodes[id][i].op],distb[id]+nodes[id][i].b);
    if(visited[nodes[id][i].op]) continue;
    visited[nodes[id][i].op] = true;
    dfsb(nodes[id][i].op);
  }
}

int main(){
  ios::sync_with_stdio(false);
  cin.tie(0);
  
  cin >> n >> m >> s >> t;
  
  REP(_,m){
    ll u,v,yen,snuk;
    cin >> u >> v >> yen >> snuk;
    nodes[u].push_back({v,yen,snuk});
    nodes[v].push_back({u,yen,snuk});
  }
  
  REP(i,n+1){
    dista[i] = distb[i] = MOD;
    visited[i] = false;
  }
  
  dista[s] = distb[t] = 0;
  
  dfsa(s);
  
  REP(i,n+1){
    visited[i] = false;
  }
  
  dfsb(t);
  
  // sからAでいける　ｔまでBでいける
  
/*  RREP(i,n+1){
    if(i-1 < 0) break;
    dista[i-1] = min(dista[i],dista[i-1]);
    distb[i-1] = min(distb[i],distb[i-1]);
  }*/
  
 /* REP(i,4){
    cout << dista[i+1] << " " << distb[i+1] << endl;
  }*/
  
  REP(i,n)
    cout << 1000000000000000 - dista[i+1] - distb[i+1] << endl;
  
  return 0;
}