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;
use std::collections::BinaryHeap;

const INF: usize = 1e16 as usize;

#[derive(Eq, PartialEq)]
struct State {
    v: usize,
    cost: usize,
}

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

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

fn main() {
    let mut sc = Scanner::new();
    let n = sc.read();
    let m = sc.read();
    let s = sc.usize_read() - 1;
    let t = sc.usize_read() - 1;

    let mut graph_a = vec![vec![]; n];
    let mut graph_b = vec![vec![]; n];
    for _ in 0..m {
        let u = sc.usize_read() - 1;
        let v = sc.usize_read() - 1;
        let a = sc.read();
        let b = sc.read();
        graph_a[u].push((v, a));
        graph_a[v].push((u, a));
        graph_b[u].push((v, b));
        graph_b[v].push((u, b));
    }

    let dist_a = dijkstra(&graph_a, s);
    let dist_b = dijkstra(&graph_b, t);

    let mut ans = vec![0; n];
    for i in (0..n).rev() {
        ans[i] = dist_a[i] + dist_b[i];
        if i < n - 1 {
            ans[i] = cmp::min(ans[i], ans[i + 1]);
        }
    }

    for &ans in &ans {
        println!("{}", (1e15 as usize) - ans);
    }
}

fn dijkstra(graph: &Vec<Vec<(usize, usize)>>, from: usize) -> Vec<usize> {
    let n = graph.len();
    let mut heap = BinaryHeap::new();
    let mut dist = vec![INF; n];
    dist[from] = 0;
    heap.push(State { v: from, cost: 0 });
    while let Some(State { v, cost: _ }) = heap.pop() {
        for &(next, weight) in &graph[v] {
            if dist[next] > dist[v] + weight {
                dist[next] = dist[v] + weight;
                heap.push(State {
                    v: next,
                    cost: dist[next],
                });
            }
        }
    }
    dist
}

struct Scanner {
    ptr: usize,
    length: usize,
    buf: Vec<u8>,
    small_cache: Vec<u8>,
}

#[allow(dead_code)]
impl Scanner {
    fn new() -> Scanner {
        Scanner {
            ptr: 0,
            length: 0,
            buf: vec![0; 1024],
            small_cache: vec![0; 1024],
        }
    }

    fn load(&mut self) {
        use std::io::Read;
        let mut s = std::io::stdin();
        self.length = s.read(&mut self.buf).unwrap();
    }

    fn byte(&mut self) -> u8 {
        if self.ptr >= self.length {
            self.ptr = 0;
            self.load();
            if self.length == 0 {
                self.buf[0] = b'\n';
                self.length = 1;
            }
        }

        self.ptr += 1;
        return self.buf[self.ptr - 1];
    }

    fn is_space(b: u8) -> bool {
        b == b'\n' || b == b'\r' || b == b'\t' || b == b' '
    }

    fn read_vec<T>(&mut self, n: usize) -> Vec<T>
    where
        T: std::str::FromStr,
        T::Err: std::fmt::Debug,
    {
        (0..n).map(|_| self.read()).collect()
    }

    fn usize_read(&mut self) -> usize {
        self.read()
    }

    fn read<T>(&mut self) -> T
    where
        T: std::str::FromStr,
        T::Err: std::fmt::Debug,
    {
        let mut b = self.byte();
        while Scanner::is_space(b) {
            b = self.byte();
        }

        for pos in 0..self.small_cache.len() {
            self.small_cache[pos] = b;
            b = self.byte();
            if Scanner::is_space(b) {
                return String::from_utf8_lossy(&self.small_cache[0..(pos + 1)])
                    .parse()
                    .unwrap();
            }
        }

        let mut v = self.small_cache.clone();
        while !Scanner::is_space(b) {
            v.push(b);
            b = self.byte();
        }
        return String::from_utf8_lossy(&v).parse().unwrap();
    }
}