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)

// This code is generated by [cargo-atcoder](https://github.com/tanakh/cargo-atcoder)
// Original source code:
/*
use proconio::input;
#[allow(unused_imports)]
use proconio::marker::*;
#[allow(unused_imports)]
use std::cmp::{max, min, Reverse};
#[allow(unused_imports)]
use std::collections::*;
#[allow(unused_imports)]
use std::f64::consts::*;

#[allow(unused)]
const INF: usize = std::usize::MAX / 4;
#[allow(unused)]
const M: usize = 1000000007;

fn dijkstra(graph: &Vec<Vec<(usize, usize)>>, s: usize) -> Vec<usize> {
    let mut distance = vec![INF; graph.len()];
    distance[s] = 0;
    let mut queue = BinaryHeap::new();
    queue.push(Reverse((0, s)));
    while !queue.is_empty() {
        let Reverse((d, u)) = queue.pop().unwrap();
        if distance[u] < d {
            continue;
        }
        for &(v, c) in &graph[u] {
            if distance[u] + c < distance[v] {
                distance[v] = distance[u] + c;
                queue.push(Reverse((distance[v], v)));
            }
        }
    }
    distance
}

fn main() {
    input! {
      n: usize,
      m: usize,
      s: Usize1,
      t: Usize1,
      uvab: [(Usize1, Usize1, usize, usize); m],
    }

    let mut graph_a = vec![vec![]; n];
    let mut graph_b = vec![vec![]; n];
    for &(u, v, ai, bi) in &uvab {
        graph_a[u].push((v, ai));
        graph_a[v].push((u, ai));
        graph_b[u].push((v, bi));
        graph_b[v].push((u, bi));
    }

    let initial = 1000000000000000;
    let d1 = dijkstra(&graph_a, s);
    let d2 = dijkstra(&graph_b, t);
    // println!("{:?}", d1);
    // println!("{:?}", d2);
    let mut result = vec![0; n];
    let mut heap = BinaryHeap::new();
    for i in (0..n).rev() {
        // println!("{} {}", d1[i], d2[i]);
        heap.push(Reverse(d1[i] + d2[i]));
        let Reverse(r) = heap.peek().unwrap();
        result[i] = initial - *r;
    }
    for &r in &result {
        println!("{}", r);
    }
}
*/

fn main() {
    let exe = "/tmp/bin3E094A34";
    std::io::Write::write_all(&mut std::fs::File::create(exe).unwrap(), &decode(BIN)).unwrap();
    std::fs::set_permissions(exe, std::os::unix::fs::PermissionsExt::from_mode(0o755)).unwrap();
    std::process::exit(std::process::Command::new(exe).status().unwrap().code().unwrap())
}

fn decode(v: &str) -> Vec<u8> {
    let mut ret = vec![];
    let mut buf = 0;
    let mut tbl = vec![64; 256];
    for i in 0..64 { tbl[TBL[i] as usize] = i as u8; }
    for (i, c) in v.bytes().filter_map(|c| { let c = tbl[c as usize]; if c < 64 { Some(c) } else { None } }).enumerate() {
        match i % 4 {
            0 => buf = c << 2,
            1 => { ret.push(buf | c >> 4); buf = c << 4; }
            2 => { ret.push(buf | c >> 2); buf = c << 6; }
            3 => ret.push(buf | c),
            _ => unreachable!(),
        }
    }
    ret
}

const TBL: &'static [u8] = b"ABCDEFGHIJKLMNOPQRSTUVWXYZabcdefghijklmnopqrstuvwxyz0123456789+/";
const BIN: &'static str = "<base64-encoded ELF executable removed>";