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::collections::BinaryHeap;
use std::cmp::Ordering;

#[derive(Eq, PartialEq, Clone, Debug)]
pub struct Rev<T>(pub T);

impl<T: PartialOrd> PartialOrd for Rev<T> {
    fn partial_cmp(&self, other: &Rev<T>) -> Option<Ordering> {
        other.0.partial_cmp(&self.0)
    }
}

impl<T: Ord> Ord for Rev<T> {
    fn cmp(&self, other: &Rev<T>) -> Ordering {
        other.0.cmp(&self.0)
    }
}

pub fn read1<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

pub fn readn<T: std::str::FromStr>(delimiter: &str) -> Vec<T> {
    let s = read1::<String>();
    s.split(delimiter).map(|e| e.parse().ok().unwrap()).collect::<Vec<T>>()
}

const INF: usize = 1_000_000_000_000_000;

fn main() {
    let nmst = readn::<usize>(" ");
    let n = nmst[0];
    let m = nmst[1];
    let s = nmst[2] - 1;
    let t = nmst[3] - 1;

    let mut yen = vec![Vec::new(); n];
    let mut snuke = vec![Vec::new(); n];

    for _ in 0..m {
        let uvab = readn::<usize>(" ");
        let u = uvab[0] - 1;
        let v = uvab[1] - 1;
        let a = uvab[2];
        let b = uvab[3];

        yen[u].push((v, a));
        yen[v].push((u, a));
        snuke[u].push((v, b));
        snuke[v].push((u, b));
    }

    let mut ds = vec![INF; n];
    let mut dt = vec![INF; n];
    ds[s] = 0;
    dt[t] = 0;

    let mut que = BinaryHeap::new();
    que.push(Rev((0, s)));
    while let Some(Rev((d, i))) = que.pop() {
        if ds[i] != d { continue; }
        for &tup in yen[i].iter() {
            let j = tup.0;
            let dd = tup.1;
            if ds[j] > d + dd {
                ds[j] = d + dd;
                que.push(Rev((d + dd, j)));
            }
        }
    }

    let mut que = BinaryHeap::new();
    que.push(Rev((0, t)));
    while let Some(Rev((d, i))) = que.pop() {
        if dt[i] != d { continue; }
        for &tup in snuke[i].iter() {
            let j = tup.0;
            let dd = tup.1;
            if dt[j] > d + dd {
                dt[j] = d + dd;
                que.push(Rev((d + dd, j)));
            }
        }
    }

    let mut que = BinaryHeap::new();
    for i in 0..n {
        let cost = ds[i] + dt[i];
        if cost > INF {
            que.push((0, i));
        } else {
            que.push((INF - cost, i));
        }
    }

    let mut di = que.pop().unwrap();
    let mut ans = String::new();
    for i in 0..n {
        while di.1 < i {
            di = que.pop().unwrap();
        }
        ans += &((di.0).to_string());
        ans += "\n";
    }
    println!("{}", ans);
}