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::io::{BufRead, BufWriter, Write};

fn main() {
    let stdin = std::io::stdin();
    let vals = stdin.lock().gets().split(' ').map(|x| x.parse().unwrap())
        .collect::<Vec<usize>>();
    let (n, m, s, t) = (vals[0], vals[1], vals[2] - 1, vals[3] - 1);
    let mut yedges = vec![Vec::new(); n];
    let mut sedges = vec![Vec::new(); n];
    for line in stdin.lock().lines().take(m).map(|x| x.unwrap()) {
        let mut it = line.split(' ').map(|x| x.parse::<usize>().unwrap());
        let (n1, n2) = (it.next().unwrap() - 1, it.next().unwrap() - 1);
        let (yen, snk) = (it.next().unwrap() as u64, it.next().unwrap() as u64);
        yedges[n1].push((n2, yen));
        yedges[n2].push((n1, yen));
        sedges[n1].push((n2, snk));
        sedges[n2].push((n1, snk));
    }
    let mut fare = vec![0; n];
    let mut d = Dijkstra::new(n, yedges, s);
    while let Some((n, c)) = d.next() {
        fare[n] = c;
    }
    let mut d = Dijkstra::new(n, sedges, t);
    while let Some((n, c)) = d.next() {
        fare[n] += c;
    }
    let mut min = std::u64::MAX;
    let mut ans = Vec::new();
    for f in fare.into_iter().rev() {
        min = std::cmp::min(min, f);
        ans.push(1_000_000_000_000_000 - min);
    }
    let stdout = std::io::stdout();
    let mut stdout = BufWriter::new(stdout.lock());
    for x in ans.into_iter().rev() {
        writeln!(stdout, "{}", x).unwrap();
    }
}

struct Dijkstra {
    edges: Vec<Vec<(usize, u64)>>,
    queue: BinHeap,
}

struct BinHeap {
    cost: Vec<u64>,
    heap: Vec<usize>,
    index: Vec<usize>,
}

impl Dijkstra {
    pub fn new(size: usize, edges: Vec<Vec<(usize, u64)>>, s: usize) -> Self {
        let mut queue = BinHeap::new(size);
        queue.set_lower_cost(s, 0);
        Dijkstra { edges: edges, queue: queue }
    }

    fn next(&mut self) -> Option<(usize, u64)> {
        self.queue.pop().map(|v| {
            for &(n, d) in &self.edges[v] {
                let c = self.queue.cost[v] + d;
                if c < self.queue.cost[n] {
                    self.queue.set_lower_cost(n, c);
                }
            }
            (v, self.queue.cost[v])
        })
    }
}

impl BinHeap {
    fn new(size: usize) -> Self {
        let cost = vec![std::u64::MAX; size];
        let heap = (0..size).collect::<Vec<_>>();
        BinHeap { cost: cost, heap: heap.clone(), index: heap }
    }

    fn set_lower_cost(&mut self, n: usize, c: u64) {
        self.cost[n] = c;
        let ni = self.index[n];
        if ni < self.heap.len() {
            self.upheap(ni);
        }
    }

    fn pop(&mut self) -> Option<usize> {
        if self.heap.len() < 2 {
            self.heap.pop()
        } else {
            let last = self.heap.len() - 1;
            self.heap.swap(0, last);
            self.index.swap(self.heap[0], self.heap[last]);
            let root = self.heap.pop();
            self.downheap(0);
            root
        }
    }

    fn upheap(&mut self, mut ni: usize) {
        let n = self.heap[ni];
        while ni > 0 {
            let pi = (ni - 1) / 2;
            if self.cost[self.heap[pi]] <= self.cost[n] {
                break;
            }
            self.index.swap(n, self.heap[pi]);
            self.heap.swap(ni, pi);
            ni = pi;
        }
    }

    fn downheap(&mut self, mut ni: usize) {
        let n = self.heap[ni];
        while self.heap.len() > ni * 2 + 1 {
            let (li, ri) = (ni * 2 + 1, ni * 2 + 2);
            let ci = if self.heap.len() <= ri ||
                self.cost[self.heap[li]] < self.cost[self.heap[ri]] {
                    li
                } else {
                    ri
                };
            if self.cost[n] <= self.cost[self.heap[ci]] {
                break;
            }
            self.index.swap(n, self.heap[ci]);
            self.heap.swap(ni, ci);
            ni = ci;
        }
    }
}

trait Gets { fn gets(&mut self) -> String; }
impl<T: std::io::BufRead> Gets for T {
    fn gets(&mut self) -> String { self.lines().next().unwrap().unwrap() } }