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)

#![allow(unused_imports)]
use std::io::{ self, Write };
use std::str::FromStr;
use std::cmp::{ min, max };
use std::collections::{ BinaryHeap, VecDeque };

#[allow(unused_macros)]
macro_rules! trace {
    ($var:expr) => {
        let _ = writeln!(&mut std::io::stderr(), ">>> {} = {:?}", stringify!($var), $var);
    };
    ($($args:expr),*) => { trace!(($($args),*)) }
}

#[allow(unused_macros)]
macro_rules! put {
    ($var:expr) => {
        let _ = writeln!(&mut std::io::stdout(), "{}", $var);
    };
    ($var:expr, $($args:expr),*) => {
        let _ = write!(&mut std::io::stdout(), "{} ", $var);
        put!($($args),*);
    };
}

const INF: u64 = 10000000000000000;

fn main() {
    let mut sc = Scanner::new();

    let n: usize = sc.cin(); // nodes
    let m: usize = sc.cin(); // edges
    let s: usize = sc.cin::<usize>() - 1;
    let t: usize = sc.cin::<usize>() - 1;

    // let mut es = vec![];
    let mut es1 = vec![vec![]; n];
    let mut es2 = vec![vec![]; n];
    for _ in 0..m {
        let u = sc.cin::<usize>() - 1;
        let v = sc.cin::<usize>() - 1;
        let a: u64 = sc.cin();
        let b: u64 = sc.cin();
        // es.push((u, v, a, b));
        es1[u].push((v, a));
        es1[v].push((u, a));
        es2[u].push((v, b));
        es2[v].push((u, b));
    }

    let mut d1 = vec![INF; n];
    let mut d2 = vec![INF; n];

    {
        d1[s] = 0;
        let mut stack = vec![s];
        while let Some(u) = stack.pop() {
            for &(v, cost) in es1[u].iter() {
                if d1[v] > d1[u] + cost {
                    d1[v] = d1[u] + cost;
                    stack.push(v);
                }
            }
        }
    }

    {
        d2[t] = 0;
        let mut stack = vec![t];
        while let Some(u) = stack.pop() {
            for &(v, cost) in es2[u].iter() {
                if d2[v] > d2[u] + cost {
                    d2[v] = d2[u] + cost;
                    stack.push(v);
                }
            }
        }
    }

    let d = (0..n).map(|i| d1[i] + d2[i]).collect::<Vec<_>>();
    // trace!(d);

    let money: u64 = 1000000000000000;
    let mut ans = vec![];
    let mut mn = d[n-1];
    for i in 0..n {
        mn = min(d[n - i - 1], mn);
        ans.push(money - mn);
    }

    for i in 0..n {
        put!(ans[n - i - 1]);
    }

}

#[allow(dead_code)]
struct Scanner { stdin: io::Stdin, buffer: VecDeque<String>, }
#[allow(dead_code)]
impl Scanner {
    fn new() -> Scanner { Scanner { stdin: io::stdin(), buffer: VecDeque::new() } }
    fn reserve(&mut self) {
        while self.buffer.len() == 0 {
            let mut line = String::new();
            let _ = self.stdin.read_line(&mut line);
            for w in line.split_whitespace() {
                self.buffer.push_back(String::from(w));
            }
        }
    }
    fn cin<T: FromStr>(&mut self) -> T {
        self.reserve();
        match self.buffer.pop_front().unwrap().parse::<T>() {
            Ok(a) => a,
            Err(_) => panic!("parse err")
        }
    }
}