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)

#[allow(unused_imports)]
use std::cmp::*;
#[allow(unused_imports)]
use std::collections::*;
use std::io::Read;
#[allow(dead_code)]
fn getline() -> String {
    let mut ret = String::new();
    std::io::stdin().read_line(&mut ret).ok().unwrap();
    ret
}
fn get_word() -> String {
    let mut stdin = std::io::stdin();
    let mut u8b: [u8; 1] = [0];
    loop {
        let mut buf: Vec<u8> = Vec::with_capacity(16);
        loop {
            let res = stdin.read(&mut u8b);
            if res.unwrap_or(0) == 0 || u8b[0] <= b' ' {
                break;
            } else {
                buf.push(u8b[0]);
            }
        }
        if buf.len() >= 1 {
            let ret = String::from_utf8(buf).unwrap();
            return ret;
        }
    }
}

#[allow(dead_code)]
fn get<T: std::str::FromStr>() -> T { get_word().parse().ok().unwrap() }

/*
 * Dijkstra's algorithm.
 * Verified by: AtCoder ABC035 (http://abc035.contest.atcoder.jp/submissions/676539)
 */

struct Dijkstra {
    edges: Vec<Vec<(usize, i64)>>, // adjacent list representation
}

/*
 * Code from https://doc.rust-lang.org/std/collections/binary_heap/
 */
#[derive(Copy, Clone, Eq, PartialEq)]
struct State {
    cost: i64,
    position: usize,
}

// The priority queue depends on `Ord`.
// Explicitly implement the trait so the queue becomes a min-heap
// instead of a max-heap.
impl Ord for State {
    fn cmp(&self, other: &State) -> Ordering {
        // Notice that the we flip the ordering here
        match other.cost.cmp(&self.cost) {
            std::cmp::Ordering::Equal => other.position.cmp(&self.position),
            x => x,
        }
    }
}

// `PartialOrd` needs to be implemented as well.
impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}


impl Dijkstra {
    fn new(n: usize) -> Self {
        Dijkstra { edges: vec![Vec::new(); n] }
    }
    fn add_edge(&mut self, from: usize, to: usize, cost: i64) {
        self.edges[from].push((to, cost));
    }
    /*
     * This function returns a Vec consisting of the distances from vertex source.
     */
    fn solve(&self, source: usize, inf: i64) -> Vec<i64> {
        let n = self.edges.len();
        let mut d = vec![inf; n];
        let mut que = std::collections::BinaryHeap::new();
        que.push(State {cost: 0, position: source});
        while let Some(State {cost, position: pos}) = que.pop() {
            if d[pos] <= cost {
                continue;
            }
            d[pos] = cost;
            for adj in &self.edges[pos] {
                que.push(State {cost: cost + adj.1, position: adj.0});
            }
        }
        return d;
    }
}

const INF: i64 = 1 << 61;


fn solve() {
    let n = get();
    let m = get();
    let s = get::<usize>() - 1;
    let t = get::<usize>() - 1;
    let mut dij_y = Dijkstra::new(n);
    let mut dij_s = Dijkstra::new(n);
    for _ in 0 .. m {
        let u = get::<usize>() - 1;
        let v = get::<usize>() - 1;
        let a = get();
        let b = get();
        dij_y.add_edge(u, v, a);
        dij_y.add_edge(v, u, a);
        dij_s.add_edge(u, v, b);
        dij_s.add_edge(v, u, b);
    }
    let sol_s = dij_y.solve(s, INF);
    let sol_t = dij_s.solve(t, INF);
    let mut ans = vec![INF; n];
    for i in 0 .. n {
        ans[i] = 1_000_000_000_000_000 - (sol_s[i] + sol_t[i]);
    }
    for i in (0 .. n - 1).rev() {
        ans[i] = max(ans[i], ans[i + 1]);
    }
    for a in ans {
        println!("{}", a);
    }
}

fn main() {
    // In order to avoid potential stack overflow, spawn a new thread.
    let stack_size = 104_857_600; // 100 MB
    let thd = std::thread::Builder::new().stack_size(stack_size);
    thd.spawn(|| solve()).unwrap().join().unwrap();
}