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: Time Limit Exceeded)

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

#[derive(Copy, Clone, Eq, PartialEq)]
struct Edge{
    from: usize,
    to: usize,
    weight: i64,
}
impl Ord for Edge {
    fn cmp(&self, other: &Edge) -> Ordering {
        other.weight.cmp(&self.weight)
    }
}
impl PartialOrd for Edge {
    fn partial_cmp(&self, other: &Edge) -> Option<Ordering> {
        Some(self.cmp(other))
    }
}


fn dijkstra_heap(graph: &Vec<Vec<Edge>>, start: usize) -> Vec<(usize, i64)> {
    let n = graph.len();
    // let mut stack: Vec<Edge>  = Vec::new();
    let mut heap = BinaryHeap::new();
    let mut dist : Vec<(usize, i64)> = vec![(std::usize::MAX, std::i64::MAX); n];
    
    heap.push((start,0));
    dist[start] = (start, 0);
    while let Some((position, cost)) = heap.pop(){

        let cost_s2f = dist[position].1;
        if cost_s2f < cost {
            continue;
        }
        for &next in graph[position].iter() {
            let cost_f2n = next.weight;
            let cost_s2n = dist[next.to].1;
            let new_edge = (next.to, cost_s2f + cost_f2n);
            // println!("!!! {}=>{} / {} {} {}",position, i.to, cost_s2n , cost_s2f , cost_f2n  );
            if cost_s2n > new_edge.1 {
                dist[next.to] = (position, cost_s2f + cost_f2n);
                heap.push(new_edge);
            }
        }
    }
    dist
}

fn dijkstra_bruteforce(graph: &Vec<Vec<Edge>>, start: usize) -> Vec<(usize, i64)> {
    let n = graph.len();
    // let mut stack: Vec<Edge>  = Vec::new();
    // let mut heap = BinaryHeap::new();
    let mut dist : Vec<(usize, i64)> = vec![(std::usize::MAX, std::i64::MAX); n];
    let mut visited : Vec<bool> = vec![false; n];

    // heap.push((start,0));
    dist[start] = (start, 0);
    loop {
        let mut position = std::usize::MAX;
        let mut cost = std::i64::MAX;
        for n in 0..n {
            if !visited[n] && dist[n].1 < cost{
                position = n;
                cost = dist[n].1;
            }
        }
        if position == std::usize::MAX {
            break;
        }

        let cost_s2f = dist[position].1;
        visited[position] = true;
        // if cost_s2f < cost {
        //     continue;
        // }
        for &next in graph[position].iter() {
            let cost_f2n = next.weight;
            let cost_s2n = dist[next.to].1;
            let new_edge = (next.to, cost_s2f + cost_f2n);
            // println!("!!! {}=>{} / {} {} {}",position, i.to, cost_s2n , cost_s2f , cost_f2n  );
            if cost_s2n > new_edge.1 {
                dist[next.to] = (position, cost_s2f + cost_f2n);
                // heap.push(new_edge);
            }
        }
    }
    dist
}


fn main() {
    let mut buf = String::new();
    std::io::stdin().read_to_string(&mut buf).unwrap();
    let mut iter = buf.split_whitespace();

    // get stdio    
    let n: usize = iter.next().unwrap().parse::<usize>().unwrap();
    let m: usize = iter.next().unwrap().parse::<usize>().unwrap();
    let s: usize = iter.next().unwrap().parse::<usize>().unwrap() - 1;
    let t: usize = iter.next().unwrap().parse::<usize>().unwrap() - 1;
    
    let edges: Vec<(usize, usize, i64, i64)> = (0..m)
        .map(|_| {
            (
                iter.next().unwrap().parse::<usize>().unwrap() - 1,
                iter.next().unwrap().parse::<usize>().unwrap() - 1,
                iter.next().unwrap().parse::<i64>().unwrap(),
                iter.next().unwrap().parse::<i64>().unwrap(),
            )
        })
        .collect();

    // construct graph
    let mut graph_a: Vec<Vec<Edge>> = (0..n)
        .map(|_| {
            Vec::new()
        }).collect();
    
    let mut graph_b: Vec<Vec<Edge>> = (0..n)
        .map(|_| {
            Vec::new()
        }).collect();

    for e in edges.iter() {
        graph_a[e.0].push( Edge{from:e.0, to:e.1, weight:e.2});
        graph_a[e.1].push( Edge{from:e.1, to:e.0, weight:e.2});
        graph_b[e.0].push( Edge{from:e.0, to:e.1, weight:e.3});
        graph_b[e.1].push( Edge{from:e.1, to:e.0, weight:e.3});
    }



    let dist_a = dijkstra_bruteforce(&graph_a, s);
    let dist_b = dijkstra_bruteforce(&graph_b, t);
    // for i in (0..n) {
    //     println!("node {} <- {} : {} / {} <- {} : {}", i, dist_a[i].0, dist_a[i].1, i, dist_b[i].0, dist_b[i].1);
    // }

    let mut dist : Vec<(i64, usize)> = dist_a.iter()
        .zip(dist_b.iter())
        .enumerate()
        .map(|(i,(a, b))| (100000_00000_00000 - (a.1 + b.1), i))
        .collect::<Vec<(i64, usize)>>();

    dist.sort_by(|a, b|b.cmp(a));

    
    let mut index:usize = 0;
    for i in 0..n {
        while dist[index].1 < i {
            index = index + 1
        }
        println!("{}", dist[index].0);
    }


    // for i in (0..n) {
    //     println!("{} : {}", dist[i].0, dist[i].1);
    // }
    // println!("{}",0);

}