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)

#[doc = " https://github.com/hatoo/competitive-rust-snippets"]
#[allow(unused_imports)]
use std::cmp::{max, min, Ordering};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::io::{stdin, stdout, BufWriter, Write};
#[allow(unused_imports)]
use std::iter::FromIterator;
#[allow(unused_macros)]
macro_rules ! debug { ( $ ( $ a : expr ) ,* ) => { eprintln ! ( concat ! ( $ ( stringify ! ( $ a ) , " = {:?}, " ) ,* ) , $ ( $ a ) ,* ) ; } }
#[macro_export]
macro_rules ! input { ( source = $ s : expr , $ ( $ r : tt ) * ) => { let mut parser = Parser :: from_str ( $ s ) ; input_inner ! { parser , $ ( $ r ) * } } ; ( parser = $ parser : ident , $ ( $ r : tt ) * ) => { input_inner ! { $ parser , $ ( $ r ) * } } ; ( new_stdin_parser = $ parser : ident , $ ( $ r : tt ) * ) => { let stdin = std :: io :: stdin ( ) ; let reader = std :: io :: BufReader :: new ( stdin . lock ( ) ) ; let mut $ parser = Parser :: new ( reader ) ; input_inner ! { $ parser , $ ( $ r ) * } } ; ( $ ( $ r : tt ) * ) => { input ! { new_stdin_parser = parser , $ ( $ r ) * } } ; }
#[macro_export]
macro_rules ! input_inner { ( $ parser : ident ) => { } ; ( $ parser : ident , ) => { } ; ( $ parser : ident , $ var : ident : $ t : tt $ ( $ r : tt ) * ) => { let $ var = read_value ! ( $ parser , $ t ) ; input_inner ! { $ parser $ ( $ r ) * } } ; }
#[macro_export]
macro_rules ! read_value { ( $ parser : ident , ( $ ( $ t : tt ) ,* ) ) => { ( $ ( read_value ! ( $ parser , $ t ) ) ,* ) } ; ( $ parser : ident , [ $ t : tt ; $ len : expr ] ) => { ( 0 ..$ len ) . map ( | _ | read_value ! ( $ parser , $ t ) ) . collect ::< Vec < _ >> ( ) } ; ( $ parser : ident , chars ) => { read_value ! ( $ parser , String ) . chars ( ) . collect ::< Vec < char >> ( ) } ; ( $ parser : ident , usize1 ) => { read_value ! ( $ parser , usize ) - 1 } ; ( $ parser : ident , $ t : ty ) => { $ parser . next ::<$ t > ( ) . expect ( "Parse error" ) } ; }
use std::io;
use std::io::BufRead;
use std::str;
pub struct Parser<R> {
    reader: R,
    buf: Vec<u8>,
    pos: usize,
}
impl Parser<io::Empty> {
    pub fn from_str(s: &str) -> Parser<io::Empty> {
        Parser {
            reader: io::empty(),
            buf: s.as_bytes().to_vec(),
            pos: 0,
        }
    }
}
impl<R: BufRead> Parser<R> {
    pub fn new(reader: R) -> Parser<R> {
        Parser {
            reader: reader,
            buf: vec![],
            pos: 0,
        }
    }
    pub fn update_buf(&mut self) {
        self.buf.clear();
        self.pos = 0;
        loop {
            let (len, complete) = {
                let buf2 = self.reader.fill_buf().unwrap();
                self.buf.extend_from_slice(buf2);
                let len = buf2.len();
                if len == 0 {
                    break;
                }
                (len, buf2[len - 1] <= 0x20)
            };
            self.reader.consume(len);
            if complete {
                break;
            }
        }
    }
    pub fn next<T: str::FromStr>(&mut self) -> Result<T, T::Err> {
        loop {
            let mut begin = self.pos;
            while begin < self.buf.len() && (self.buf[begin] <= 0x20) {
                begin += 1;
            }
            let mut end = begin;
            while end < self.buf.len() && (self.buf[end] > 0x20) {
                end += 1;
            }
            if begin != self.buf.len() {
                self.pos = end;
                return str::from_utf8(&self.buf[begin..end]).unwrap().parse::<T>();
            } else {
                self.update_buf();
            }
        }
    }
}
#[allow(unused_macros)]
macro_rules ! debug { ( $ ( $ a : expr ) ,* ) => { eprintln ! ( concat ! ( $ ( stringify ! ( $ a ) , " = {:?}, " ) ,* ) , $ ( $ a ) ,* ) ; } }
#[doc = " https://github.com/hatoo/competitive-rust-snippets"]
const BIG_STACK_SIZE: bool = true;
#[allow(dead_code)]
fn main() {
    use std::thread;
    if BIG_STACK_SIZE {
        thread::Builder::new()
            .stack_size(32 * 1024 * 1024)
            .name("solve".into())
            .spawn(solve)
            .unwrap()
            .join()
            .unwrap();
    } else {
        solve();
    }
}
fn solve() {
    input!{
        n:usize,m:usize,s:usize,t:usize,
        uvab:[(usize,usize,i64,i64);m],
    }

    let mut g=vec![vec![];n+1];
    let mut rg=vec![vec![];n+1];
    for (u,v,a,b) in uvab {
        g[u].push(Edge{to:v,cost:a});
        g[v].push(Edge{to:u,cost:a});
        
        rg[u].push(Edge{to:v,cost:b});
        rg[v].push(Edge{to:u,cost:b});
    }
    // 円だけを使ってsからいく最小コスト
    let dp=dijkstra_heap(&g,s,std::i64::MAX);
    // スヌークだけを使ってtに向かう最小コスト
    let rdp=dijkstra_heap(&rg,t,std::i64::MAX);

    let mut mincost=std::i64::MAX;
    let mut ans=vec![];
    for i in (1..n+1).rev() {
        // 新しく開かれる両替所を使った時のコスト
        let cost=dp[i]+rdp[i];
        mincost=min(mincost,cost);
        ans.push(1_000_000_000_000_000-mincost);
    }
    ans.reverse();
    for x in ans {
        println!("{}",x);
    }
}
    #[derive(Clone,Copy,Debug)]
    pub struct Edge {
        to: usize,
        cost: i64,
    }

    pub fn dijkstra_heap(g: &[Vec<Edge>], s: usize, inf: i64) -> Vec<i64> {
        let n = g.len();
        let mut queue = std::collections::BinaryHeap::new(); // max-heap
        let mut d = vec![inf; n];

        d[s] = 0;
        queue.push((0, s));

        while let Some((cost, v)) = queue.pop() {
            let cost = -cost;

            if d[v] < cost {
                continue;
            }
            for e in &g[v] {
                let new_cost = cost + e.cost;
                if d[e.to] > new_cost {
                    d[e.to] = new_cost;
                    queue.push((-new_cost, e.to));
                }
            }
        }

        d
    }