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: Wrong Answer)

use std::collections::{BinaryHeap, HashSet};
use std::io::{stdin, Read};

macro_rules! parse {
    ($iter: ident) => {
        $iter.next().unwrap().parse().unwrap()
    };
    ($iter: ident,2) => {
        (parse!($iter), parse!($iter))
    };
    ($iter: ident,3) => {
        (parse!($iter), parse!($iter), parse!($iter))
    };
    ($iter: ident,4) => {
        (parse!($iter), parse!($iter), parse!($iter), parse!($iter))
    };
    ($iter: ident, $n: ident) => {
        $iter.take($n).map(|e| e.parse().unwrap()).collect();
    };
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    let mut iter = s.split_whitespace();
    let (n, m, s, t): (usize, usize, usize, usize) = parse!(iter, 4);
    let mut ga = vec![vec![]; n];
    let mut gb = vec![vec![]; n];
    for _ in 0..m {
        let (u, v, a, b): (usize, usize, isize, isize) = parse!(iter, 4);
        ga[u - 1].push((v - 1, a));
        ga[v - 1].push((u - 1, a));
        gb[u - 1].push((v - 1, b));
        gb[v - 1].push((u - 1, b));
    }
    let da = get_cost(n, s, ga);
    let db = get_cost(n, t, gb);
    let mut d: BinaryHeap<(isize, usize)> = (0..n).map(|i| (-da[i] - db[i], i)).collect();
    let (mut cost, mut curi) = d.pop().unwrap();
    for i in 0..n {
        while curi < i {
            let new = d.pop().unwrap();
            cost = new.0;
            curi = new.1;
        }
        println!("{}", 1000000000000000 + cost); // DEBUG
    }
}

fn get_cost(n: usize, start: usize, g: Vec<Vec<(usize, isize)>>) -> Vec<isize> {
    let mut da = vec![1000000000; n];
    da[start - 1] = 0;
    let mut q: BinaryHeap<(isize, usize)> = (0..n).map(|i| (-(da[i] as isize), i)).collect();
    let mut seen = HashSet::new();
    while q.len() > 0 {
        let (uw, u) = q.pop().unwrap();
        if seen.contains(&u) {
            continue;
        }
        seen.insert(u);
        for &(v, vw) in &g[u] {
            let alt = -uw + vw;
            if da[v] > alt {
                da[v] = alt;
                q.push((-alt, v));
            }
        }
    }
    da
}