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: Time Limit Exceeded)

use std::collections::*;
use std::io::*;
use std::str::FromStr;
use utils::*;

pub fn main() {
    let i = stdin();
    let mut o = Vec::new();
    run(i.lock(), &mut o);
    stdout().write_all(&o).unwrap();
}

fn run<R: BufRead, W: Write>(i: R, o: &mut W) {
    let mut i = CpReader::new(i);
    let (n, m, s, t) = i.read4::<usize, usize, usize, usize>();
    let mut ls = vec![Vec::new(); n * 2];
    for _ in 0..m {
        let (u, v, a, b) = i.read4::<usize, usize, usize, usize>();
        let (u, v) = (u - 1, v - 1);

        ls[u].push((v, a));
        ls[v].push((u, a));

        let (u, v) = (u + n, v + n);
        ls[u].push((v, b));
        ls[v].push((u, b));
    }
    let rs = solve(n, s - 1, t - 1, ls);
    for r in rs {
        writeln!(o, "{}", 1000000000000000 - r).unwrap();
    }
}
fn solve(n: usize, s: usize, t: usize, ls: Vec<Vec<(usize, usize)>>) -> Vec<usize> {
    let mut results = Vec::new();
    let mut q = BinaryHeap::new();
    let mut costs = vec![usize::max_value(); n * 2];
    costs[s] = 0;
    for y in (0..n + 1).rev() {
        if y < n {
            if costs[y + n] > costs[y] {
                costs[y + n] = costs[y];
                q.push((y + n, costs[y]));
            }
        } else {
            q.push((s, 0));
        }
        while let Some((i0, c0)) = q.pop() {
            for &(i1, c) in &ls[i0] {
                let c1_new = c0 + c;
                let c1_old = costs[i1];
                if c1_new < c1_old {
                    costs[i1] = c1_new;
                    if i1 != t + n {
                        q.push((i1, c1_new));
                    }
                }
            }
        }
        if y < n {
            results.push(costs[t + n]);
        }
    }
    results.reverse();
    results
}

mod utils {
    use super::*;

    pub struct CpReader<R: BufRead> {
        r: R,
        s: String,
    }

    macro_rules! fn_read {
        {$f:ident($($v:ident: $t:ident),*)} => {
            pub fn $f<$($t: FromStr),*>(&mut self) -> ($($t),*) {
                let i = &mut self.read_line().split(' ');
                $(
                    let $v = i.next().unwrap().parse().ok().unwrap();
                )*
                ($($v),*)
            }
        };
    }

    impl<R: BufRead> CpReader<R> {
        pub fn new(r: R) -> Self {
            CpReader {
                r: r,
                s: String::new(),
            }
        }
        pub fn read_line(&mut self) -> &str {
            self.s.clear();
            self.r.read_line(&mut self.s).unwrap();
            self.s.trim()
        }

        fn_read! { read4(v1: T1, v2: T2, v3: T3, v4: T4) }
    }
}