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: Wrong Answer)

use std::io::{stdin, Read};

use std::cmp::Ordering;
use std::cmp::Ordering::*;
use std::cmp::PartialOrd;

#[derive(PartialEq, Eq)]
struct Rev<T>(T);

impl<T: Ord> PartialOrd for Rev<T> {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        Some(match self.0.cmp(&other.0) {
            Less => Greater,
            Equal => Equal,
            Greater => Less,
        })
    }
}

impl<T: Ord> Ord for Rev<T> {
    fn cmp(&self, other: &Self) -> Ordering {
        self.partial_cmp(other).unwrap()
    }
}

#[allow(unused_macros)]
macro_rules! parse {
    ($it: ident ) => {};
    ($it: ident, ) => {};
    ($it: ident, $var:ident : $t:tt $($r:tt)*) => {
        let $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
    ($it: ident, mut $var:ident : $t:tt $($r:tt)*) => {
        let mut $var = parse_val!($it, $t);
        parse!($it $($r)*);
    };
}

#[allow(unused_macros)]
macro_rules! parse_val {
    ($it: ident, [$t:tt; $len:expr]) => {
        (0..$len).map(|_| parse_val!($it, $t)).collect::<Vec<_>>();
    };
    ($it: ident, ($($t: ty),*)) => {
        ($(parse_val!($it, $t)),*)
    };
    ($it: ident, $t: ty) => {
        $it.next().unwrap().parse::<$t>().unwrap()
    };
}

fn solve(s: &str) {
    let mut it = s.split_whitespace();
    parse!(it, n: usize, m: usize, s: usize, t: usize);
    let mut g = vec![vec![]; n];
    for _ in 0..m {
        parse!(it, u: usize, v: usize, a: usize, b: usize);
        g[u - 1].push((v - 1, a, b));
        g[v - 1].push((u - 1, a, b));
    }
    let mut q = std::collections::BinaryHeap::new();
    let mut froms = vec![None; n];
    froms[s - 1] = Some(0);
    q.push(Rev((0, s - 1)));
    while let Some(Rev((l, u))) = q.pop() {
        froms[u] = Some(l);
        for &(v, a, _) in g[u].iter() {
            if froms[v] != None {
                continue;
            }
            q.push(Rev((l + a, v)));
        }
    }

    let mut q = std::collections::BinaryHeap::new();
    let mut fromt = vec![None; n];
    fromt[t - 1] = Some(0);
    q.push(Rev((0, t - 1)));
    while let Some(Rev((l, u))) = q.pop() {
        fromt[u] = Some(l);
        for &(v, _, b) in g[u].iter() {
            if fromt[v] != None {
                continue;
            }
            q.push(Rev((l + b, v)));
        }
    }

    let mut costs = (0..n)
        .map(|i| Rev((froms[i].unwrap() + fromt[i].unwrap(), i)))
        .collect::<std::collections::BinaryHeap<_>>();
    let money = 1_000_000_000_000_000usize;
    for i in 0..n {
        loop {
            let &Rev((cost, u)) = costs.peek().unwrap();
            if u <= i {
                costs.pop();
            }
            println!("{}", money - cost);
            break;
        }
    }
}

fn main() {
    let mut s = String::new();
    stdin().read_to_string(&mut s).unwrap();
    solve(&s);
}

#[cfg(test)]
mod tests {
    use super::*;
    #[test]
    fn test_input() {
        let s = "
        4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1
        ";
        solve(s);
    }
}