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::cmp::Ordering;
use std::collections::BinaryHeap;
const INF:i64 = 100000_00000;

#[derive(PartialEq, Debug)]
struct MinInt {
    value:i64,
}

impl Eq for MinInt {}

impl PartialOrd for MinInt {
    fn partial_cmp(&self, other: &Self) -> Option<Ordering> {
        other.value.partial_cmp(&self.value)
    }
}

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

fn make_pair(x: i64, y:usize) -> (MinInt, usize) {
    (MinInt{value: x}, y)
}

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

fn solve(edges: &Vec<Vec<Edge>>, start_idx: usize, num_apexes: usize) -> Vec<i64> {
    let mut d: Vec<i64> = vec![INF; num_apexes];
    d[start_idx] = 0;
    let mut que = BinaryHeap::new();
    que.push(make_pair(0, start_idx));

    while let Some((u, v)) = que.pop() {
        if d[v] < u.value {
            continue
        }
        for e in &edges[v] {
            if d[e.to] > d[v] + e.cost {
                d[e.to] = d[v] + e.cost;
                que.push(make_pair(d[e.to], e.to));
            }
        }
    }
    d
}


fn read<T: std::str::FromStr>() -> T {
    let mut s = String::new();
    std::io::stdin().read_line(&mut s).ok();
    s.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
    read::<String>().split_whitespace()
        .map(|e| e.parse().ok().unwrap()).collect()
}

fn main() {
    let v:Vec<i64> = read_vec();
    let (n, m, s, t) = (v[0], v[1], v[2] - 1, v[3] - 1);
    let mut edges:Vec<Vec<Edge>> = vec![Vec::new(); 2 * m as usize];
    let mut sunuke_edges:Vec<Vec<Edge>> = vec![Vec::new(); 2 * m as usize];

    for i in 0..m {
        let vv:Vec<i64> = read_vec();
        let (u1, v1, a1, b1) = (vv[0] - 1, vv[1] - 1, vv[2], vv[3]);
        edges[v1 as usize].push(Edge{to: u1 as usize, cost: a1});
        edges[u1 as usize].push(Edge{to: v1 as usize, cost: a1});

        sunuke_edges[u1 as usize].push(Edge{to: v1 as usize, cost: b1});
        sunuke_edges[v1 as usize].push(Edge{to: u1 as usize, cost: b1});
    }
    let s = s as usize;
    let t = t as usize;
    let n = n as usize;
    let yen_costs = solve(&edges, s, n);
    println!("{:?}", yen_costs);
    let sunuke_costs = solve(&sunuke_edges, t, n);
    println!("{:?}", sunuke_costs);
    let mut all_costs:Vec<i64> = vec![0; n];


    for i in 0..n {
        all_costs[i] = yen_costs[i] + sunuke_costs[i];
    }
    println!("{:?}", all_costs);

    let mut min_cost = INF;
    let mut min_costs:Vec<i64> = vec![0; n];
    for i in 0..n {
        min_cost = std::cmp::min(all_costs[n - 1 - i], min_cost);
        min_costs[n - i - 1] = min_cost;
    }
    println!("{:?}", min_costs);
    for i in 0..n {
        println!("{:?}", 1000000000000000 - min_costs[i]);
    }
}