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::collections::BinaryHeap;

fn read<T: std::str::FromStr>() -> T {
    let stdin = stdin();
    let mut buf = String::new();
	stdin.lock().read_line(&mut buf);
	buf.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().trim().split_whitespace()
        .map(|w| w.parse().ok().unwrap()).collect()
}

fn read_mat<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

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

impl Ord for State {
    fn cmp(&self, other: &State) -> std::cmp::Ordering {
        if self.cost == other.cost {
            self.pos.cmp(&other.pos)
        } else {
            other.cost.cmp(&self.cost)
        }
    }
}
impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<std::cmp::Ordering> {
        Some(self.cmp(other))
    }
}

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

fn dijkstra_all(mat: &Vec<Vec<Edge>>, s: usize) -> Vec<usize> {
    let mut heap = BinaryHeap::new();
    let mut cost: Vec<usize> = vec![usize::max_value();mat.len()];
    cost[s] = 0;
    heap.push(State {cost: 0, pos: s});
    while let Some(State {cost: c, pos: p}) = heap.pop() {
        if c > cost[p] {
            continue;
        }
        for e in &mat[p] {
            let next = State {cost: c + e.cost, pos : e.node};
            if next.cost < cost[next.pos] {
                heap.push(next);
                cost[next.pos] = next.cost;
            }
        }
    }
    cost
}

fn main() {
    let v = read_vec::<usize>();
    let (n,m,s,t) = (v[0],v[1],v[2]-1,v[3]-1);
    let w = read_mat::<usize>(m);
    let mut cy: Vec<Vec<Edge>> = vec![Vec::new();n];
    let mut cs: Vec<Vec<Edge>> = vec![Vec::new();n];
    for x in w {
        let u = x[0]-1;
        let v = x[1]-1;
        cy[u].push(Edge {node: v, cost: x[2]});
        cy[v].push(Edge {node: u, cost: x[2]});
        cs[u].push(Edge {node: v, cost: x[3]});
        cs[v].push(Edge {node: u, cost: x[3]});
    }
    let route1 = dijkstra_all(&cy, s);
    let route2 = dijkstra_all(&cs, t);
    let mut route_tot: Vec<(usize,usize)> = (0..n).map(|i| (route1[i] + route2[i], i)).collect();
    route_tot.sort_by(|a,b| a.0.cmp(&b.0));
    let mut idx = 0;
    for y in 0..n {
        while route_tot[idx].1 < y {
            idx += 1;
        }
        println!("{}",10usize.pow(15) - route_tot[idx].0);
    }
}