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)

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

#[derive(Copy, Clone, Eq, PartialEq)]
struct State {
    cost: i64,
    position: usize,
}

impl Ord for State {
    fn cmp(&self, other: &State) -> Ordering{
        other.cost.cmp(&self.cost)
    }
}


impl PartialOrd for State {
    fn partial_cmp(&self, other: &State) -> Option<Ordering>{
        Some(self.cmp(other))
    }
}

#[derive(Clone)]
struct Edge {
    node: usize,
    cost: i64,
}

fn dijkstra(adj_list: &Vec<Vec<Edge>>, start: usize) -> Vec<i64>{
    let mut dist: Vec<_> = (0..adj_list.len()).map(|_| i64::max_value()).collect();

    let mut heap = BinaryHeap::new();

    dist[start] = 0;
    heap.push(State {
        cost: 0,
        position: start,
    });

    while let Some(State {cost, position}) = heap.pop() {
        if cost > dist[position]{
            continue;
        }
        for edge in &adj_list[position] {
            let next = State{
                cost: cost + edge.cost,
                position: edge.node,
            };
            if next.cost < dist[next.position]{
                heap.push(next);
                dist[next.position] = next.cost;
            }
        }
    }

    dist
     
}



fn main() {
    let s = std::io::stdin();
    let mut sc = Scanner { stdin: s.lock() };
    let n:usize = sc.read();
    let m:usize = sc.read();
    let mut s:usize = sc.read();
    let mut t:usize = sc.read();
    s -= 1; t -= 1;
    let mut yen_graph = vec![vec![];n];
    let mut snuuk_graph = vec![vec![];n];
    for _ in 0..m{
        let mut u:usize = sc.read();
        let mut v:usize = sc.read();
        let a:i64 = sc.read();
        let b:i64 = sc.read();
        u -= 1; v -= 1;
        yen_graph[u].push(Edge{
            node: v,
            cost: a,
        });
        yen_graph[v].push(Edge{
            node: u,
            cost: a,
        });
        snuuk_graph[u].push(Edge{
            node: v,
            cost: b,
        });
        snuuk_graph[v].push(Edge{
            node: u,
            cost: b,
        });
    }
    
    let yen_shortest_path: Vec<_> = dijkstra(&yen_graph, s);
    let snuuk_shortest_path: Vec<_> = dijkstra(&snuuk_graph, t);
    let mut sum_shortest_path = Vec::new();
    for i in 0..n {
        sum_shortest_path.push((i, yen_shortest_path[i] + snuuk_shortest_path[i]));
    }
    sum_shortest_path.sort_by(|a, b| (a.1).cmp(&b.1));
    let mut years: usize = 0;
    let mut ans: Vec<i64> = Vec::new();
    for i in 0..n{
        while sum_shortest_path[years].0 < i{
            years += 1;
        }
        while sum_shortest_path[years].0 < i{
            years += 1;
        }

        ans.push(sum_shortest_path[years].1);
    }
    let start_money = 1e15 as i64;
    for i in 0..n {
        println!("{}", start_money - ans[i]);
    }

    
    
}

pub struct Scanner<R> {
    stdin: R,
}

impl<R: std::io::Read> Scanner<R> {
    pub fn read<T: std::str::FromStr>(&mut self) -> T {
        use std::io::Read;
        let buf = self
            .stdin
            .by_ref()
            .bytes()
            .map(|b| b.unwrap())
            .skip_while(|&b| b == b' ' || b == b'\n')
            .take_while(|&b| b != b' ' && b != b'\n')
            .collect::<Vec<_>>();
        unsafe { std::str::from_utf8_unchecked(&buf) }
        .parse()
            .ok()
            .expect("Parse error.")
    }
    pub fn vec<T: std::str::FromStr>(&mut self, n: usize) -> Vec<T> {
        (0..n).map(|_| self.read()).collect()
    }
    pub fn chars(&mut self) -> Vec<char> {
        self.read::<String>().chars().collect()
    }
}