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


fn main() {
    let (n, m, s, t) = {
        let i = read::<usize>();
        (i[0], i[1], i[2] - 1, i[3] - 1)
    };

    let mut vec_s = vec![Vec::new(); n];
    let mut vec_t = vec![Vec::new(); n];

    for _ in 0..m {
        let (u, v, a, b) = {
            let i = read::<usize>();
            (i[0] - 1, i[1] - 1, i[2], i[3])
        };

        vec_s[u].push(Edge { node: v, cost: a });
        vec_s[v].push(Edge { node: u, cost: a });
        vec_t[u].push(Edge { node: v, cost: b });
        vec_t[v].push(Edge { node: u, cost: b });
    }

    let dist_s = dijkstra(&vec_s, s);
    let dist_t = dijkstra(&vec_t, t);

    let mut ans = vec![::std::usize::MAX; n];
    for i in (0..n).rev() {
        ans[i] = dist_s[i] + dist_t[i];
        if i + 1 < n {
            ans[i] = ::std::cmp::min(ans[i], ans[i + 1]);
        }
    }
    for a in ans {
        println!("{}", 1_000_000_000_000_000 - a);
    }
}

#[derive(Debug, Copy, Clone, Eq, PartialEq)]
struct State {
    cost: usize,
    pos: usize,
}

impl Ord for State {
    fn cmp(&self, other: &State) -> Ordering {
        match other.cost.cmp(&self.cost) {
            Ordering::Less => Ordering::Less,
            Ordering::Greater => Ordering::Greater,
            Ordering::Equal => {
                self.pos.cmp(&other.pos)
            }
        }
    }
}

impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}


#[derive(Debug, Copy, Clone, Eq, PartialEq)]
struct Edge {
    node: usize,
    cost: usize,
}

fn dijkstra(adj: &Vec<Vec<Edge>>, s: usize) -> Vec<usize> {
    let mut dist = (0..adj.len()).map(|_| ::std::usize::MAX)
        .collect::<Vec<usize>>();

    let mut q = BinaryHeap::new();
    dist[s] = 0;
    q.push( State { cost: 0, pos: s });

    while let Some( State { cost, pos }) = q.pop() {

        if cost > dist[pos] { continue; }

        for e in &adj[pos] {
            let s = State { cost: cost + e.cost, pos: e.node };
            if s.cost < dist[s.pos] {
                q.push(s);
                dist[s.pos] = s.cost;
            }
        }
    }

    dist
}


#[allow(dead_code)]
fn read<T>() -> Vec<T>
where T:
    std::str::FromStr,
    T::Err: std::fmt::Debug {

    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.split_whitespace()
        .map(|s| s.trim().parse().unwrap())
        .collect()
}

#[allow(dead_code)]
fn read_one<T>() -> T
where T:
    std::str::FromStr,
    T::Err: std::fmt::Debug {

    let mut buf = String::new();
    io::stdin().read_line(&mut buf).unwrap();
    buf.trim().parse().unwrap()
}