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 io::*;
use std::cmp::{max, min};
use std::collections::BinaryHeap;

fn main() {
    let con = read_string();
    let mut sc = Scanner::new(&con);

    let n: usize = sc.next();
    let m: usize = sc.next();
    let mut s: usize = sc.next();
    let mut t: usize = sc.next();
    s -= 1;
    t -= 1;

    let mut ge: Vec<Vec<(usize, i64)>> = vec![vec![]; n];
    let mut gs: Vec<Vec<(usize, i64)>> = vec![vec![]; n];

    for _ in 0..m {
        let mut u: usize = sc.next();
        let mut v: usize = sc.next();
        let a: i64 = sc.next();
        let b: i64 = sc.next();

        u -= 1;
        v -= 1;

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

        gs[u].push((v, b));
        gs[v].push((u, b));
    }

    let de = dist(&ge, s);
    let ds = dist(&gs, t);

    let mut maxv = 0;
    let mut ans = vec![];

    for i in (0..n).rev() {
        maxv = max(maxv, 1000_000_000_000_000 - de[i] - ds[i]);
        ans.push(maxv);
    }

    for &a in ans.iter().rev() {
        println!("{}", a);
    }
}

fn dist(g: &Vec<Vec<(usize, i64)>>, s: usize) -> Vec<i64> {
    let n = g.len();
    let mut dist = vec![-1; n];

    let mut q = BinaryHeap::<(i64, usize)>::new();
    q.push((0, s));

    while let Some((dep, cur)) = q.pop() {
        let dep = -dep;
        if dist[cur] >= 0 {
            continue;
        }

        dist[cur] = dep;

        for &(next, w) in g[cur].iter() {
            if dist[next] >= 0 {
                continue;
            }
            q.push((-(dep + w), next));
        }
    }

    dist
}

//-----

pub mod io {
    use std;
    use std::str::FromStr;

    pub struct Scanner<'a> {
        iter: std::str::SplitWhitespace<'a>,
    }

    impl<'a> Scanner<'a> {
        pub fn new(s: &'a str) -> Scanner<'a> {
            Scanner {
                iter: s.split_whitespace(),
            }
        }

        pub fn next<T: FromStr>(&mut self) -> T {
            let s = self.iter.next().unwrap();
            if let Ok(v) = s.parse::<T>() {
                v
            } else {
                panic!("Parse error")
            }
        }

        pub fn next_vec_len<T: FromStr>(&mut self) -> Vec<T> {
            let n: usize = self.next();
            self.next_vec(n)
        }

        pub fn next_vec<T: FromStr>(&mut self, n: usize) -> Vec<T> {
            (0..n).map(|_| self.next()).collect()
        }
    }

    pub fn read_string() -> String {
        use std::io::Read;

        let mut s = String::new();
        std::io::stdin().read_to_string(&mut s).unwrap();
        s
    }

    pub fn read_line() -> String {
        let mut s = String::new();
        std::io::stdin().read_line(&mut s).unwrap();
        s.trim_right().to_owned()
    }
}