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)

use std::io::{stdin, Read, StdinLock};
use std::str::FromStr;
use std::collections::HashMap;
use std::collections::BinaryHeap;
use std::cmp;


struct Scanner<'a> {
    cin: StdinLock<'a>,
}
 
impl<'a> Scanner<'a> {
    fn new(cin: StdinLock<'a>) -> Scanner<'a> {
        Scanner { cin: cin }
    }
 
    fn read1<T: FromStr>(&mut self) -> Option<T> {
        let token = self.cin.by_ref().bytes().map(|c| c.unwrap() as char)
            .skip_while(|c| c.is_whitespace())
            .take_while(|c| !c.is_whitespace())
            .collect::<String>();
        token.parse::<T>().ok()
    }
 
    fn read<T: FromStr>(&mut self) -> T {
        self.read1().unwrap()
    }
}

fn d(g: &HashMap<usize, Vec<(usize, (usize, usize))>>, ck: usize, s: usize, n: usize) -> Vec<usize> {
    let mut pq = BinaryHeap::new();
    let mut dist: Vec<usize> = (0..(n+1)).map(|_| usize::max_value()).collect();
    dist[s] = 0;
    pq.push((usize::max_value(), s));
    while let Some((c, u)) = pq.pop() {
        // println!("{}", pq.len());
        if dist[u] > usize::max_value() - c { continue; }
        for &(v, cost) in &g[&u] {
            let alt = dist[u] + if ck == 0 {cost.0} else {cost.1};
            if dist[v] > alt {
                pq.push((usize::max_value()-alt, v));
                dist[v] = alt;
            }
        }
    }

    dist
}

fn main(){
    let cin = stdin();
    let cin = cin.lock();
    let mut sc = Scanner::new(cin);  // スキャナの定義

    let n:usize = sc.read();
    let m:usize = sc.read();
    let s:usize = sc.read();
    let t:usize = sc.read();

    let mut g:HashMap<usize, Vec<(usize, (usize, usize))>> = HashMap::new();

    for i in 1..(n+1){
        g.insert(i, Vec::new());
    }
    for _ in 0..m{    
        let u:usize = sc.read();
        let v:usize = sc.read();
        let a:usize = sc.read();
        let b:usize = sc.read();
        // println!("{:?}",g);  
        // (*g.get_mut(&u).unwrap_or(&mut Vec::new())).push((v, (a, b)));
        // (*g.get_mut(&v).unwrap_or(&mut Vec::new())).push((u, (a, b)));
        (*g.get_mut(&u).unwrap()).push((v, (a, b)));
        (*g.get_mut(&v).unwrap()).push((u, (a, b)));
    }

    let dist0 = d(&g, 0, s, n);
    let dist1 = d(&g, 1, t, n);
    let mut ans: Vec<usize> = (0..(n+1)).map(|_| usize::max_value()).collect();
    for i in (0..(n)).rev(){
        ans[i] = cmp::min(ans[i+1], dist1[i+1] + dist0[i+1]);
    }
    for i in 0..n{
        println!("{}", 1e+15 as usize - ans[i]);
    }
    
}