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 Rust (1.15.1) (judge verdict: Accepted)

#![allow(non_snake_case,unused_imports)]
use std::io::Read;
use std::cmp::max;
use std::cmp::min;
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::HashSet;
use std::collections::BinaryHeap;

fn Dijkstra_dist(N:usize,E:&Vec<HashMap<usize,usize>>,start:usize) -> Vec<isize>{
    let mut D = (0..N).map(|_| -1).collect::<Vec<_>>();
    let mut hq = BinaryHeap::new();
    hq.push((0,start));
    while hq.len()>0{
        let (d0,i) = hq.pop().unwrap();
        let d0 = -d0;
        if D[i] != -1 {continue;}
        D[i] = d0;
        for (&j,&c) in E[i].iter(){
            if D[j] != -1 {continue;}
            hq.push((-(d0+c as isize),j));
        }
    }
    D
}

fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();
    let N = iter.next().unwrap().parse::<usize>().unwrap();
    let M = iter.next().unwrap().parse::<usize>().unwrap();
    let s = iter.next().unwrap().parse::<usize>().unwrap()-1;
    let t = iter.next().unwrap().parse::<usize>().unwrap()-1;
    let UVAB = (0..M).map(|_| (
        iter.next().unwrap().parse::<usize>().unwrap()-1,
        iter.next().unwrap().parse::<usize>().unwrap()-1,
        iter.next().unwrap().parse::<usize>().unwrap(),
        iter.next().unwrap().parse::<usize>().unwrap()
    )
    ).collect::<Vec<_>>();

    let mut Ea = (0..N).map(|_| HashMap::new()).collect::<Vec<_>>();
    let mut Eb = (0..N).map(|_| HashMap::new()).collect::<Vec<_>>();
    for &(u,v,a,b) in &UVAB{
        Ea[u].insert(v,a);
        Ea[v].insert(u,a);
        Eb[u].insert(v,b);
        Eb[v].insert(u,b);
    }

    let Da = Dijkstra_dist(N,&Ea,s);
    let Db = Dijkstra_dist(N,&Eb,t);

    let mut ans = (0..N).map(|_| 0).collect::<Vec<_>>();
    ans[N-1] = Da[N-1]+Db[N-1];
    for i in 0..N-1{
        let i = N-2-i;
        ans[i] = min(ans[i+1], Da[i]+Db[i]);
    }
    for i in 0..N{
        println!("{}",1e15 as isize - ans[i]);
    }
}