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)

use std::cmp::{ min, max };
use std::collections::{ BinaryHeap, VecDeque };

struct GraphDijkstra {
    edges: Vec<Vec<(usize, i64)>>, // adjacent list
    n: usize
}

impl GraphDijkstra {
    fn new(n: usize) -> Self {
        GraphDijkstra {
            edges: vec![Vec::new(); n],
            n: n
        }
    }

    fn add_costs(&mut self, from: usize, to: usize, cost: i64, directed: bool) {
        if directed {
            self.edges[from].push((to, cost));
        } else {
            self.edges[from].push((to, cost));
            self.edges[to].push((from, cost));
        }

    }

    fn dijkstra_search(&self, start: usize, inf: i64) -> Vec<i64> {
        let mut dist: Vec<i64> = vec![inf; self.n];
        dist[start] = 0;
        let mut heap = BinaryHeap::new();
        heap.push((0, start));
        while let Some((_, v)) = heap.pop() {
            for edge in &self.edges[v] {
                let (u, cost) = *edge;
                if dist[u] > dist[v] + cost {
                    dist[u] = dist[v] + cost;
                    heap.push(( -dist[u] , u ));
                    //println!("{}, {}", u, v)
                }
            }
        }
        return dist
    }
}


macro_rules! input {
    (source = $s:expr, $($r:tt)*) => {
        let mut iter = $s.split_whitespace();
        let mut next = || { iter.next().unwrap() };
        input_inner!{next, $($r)*}
    };
    ($($r:tt)*) => {
        let stdin = std::io::stdin();
        let mut bytes = std::io::Read::bytes(std::io::BufReader::new(stdin.lock()));
        let mut next = move || -> String{
            bytes
                .by_ref()
                .map(|r|r.unwrap() as char)
                .skip_while(|c|c.is_whitespace())
                .take_while(|c|!c.is_whitespace())
                .collect()
        };
        input_inner!{next, $($r)*}
    };
}

macro_rules! input_inner {
    ($next:expr) => {};
    ($next:expr, ) => {};

    ($next:expr, $var:ident : $t:tt $($r:tt)*) => {
        let $var = read_value!($next, $t);
        input_inner!{$next $($r)*}
    };
}

macro_rules! read_value {
    ($next:expr, ( $($t:tt),* )) => {
        ( $(read_value!($next, $t)),* )
    };

    ($next:expr, [ $t:tt ; $len:expr ]) => {
        (0..$len).map(|_| read_value!($next, $t)).collect::<Vec<_>>()
    };

    ($next:expr, chars) => {
        read_value!($next, String).chars().collect::<Vec<char>>()
    };

    ($next:expr, usize1) => {
        read_value!($next, usize) - 1
    };

    ($next:expr, $t:ty) => {
        $next().parse::<$t>().expect("Parse error")
    };
}

fn main() {
    input! {
    n: usize,
    m: i32,
    s: usize,
    t: usize,
    }
    let mut graph_a = GraphDijkstra::new(n);
    let mut graph_b = GraphDijkstra::new(n);

    for _ in 0..m {
        input! {
        u: usize,
        v: usize,
        a: i64,
        b:i64,
        }
        graph_a.add_costs(u-1, v-1, a, false);
        graph_b.add_costs(u-1, v-1, b, false);
    }

    let inf: i64 = 1 << 61;
    let result_a = graph_a.dijkstra_search(s-1, inf);
    let result_b = graph_b.dijkstra_search(t-1, inf);
    let mut ans_list: Vec<i64> = vec![inf; n];
    for i in 0..n {
        ans_list[i] = 1_000_000_000_000_000 - (result_a[i] + result_b[i]);
    }
    for i in (0..n-1).rev() {
        ans_list[i] = max(ans_list[i], ans_list[i+1]);
    }
    for i in 0..n {
        println!("{}", ans_list[i])
    }
}