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 (1.42.0) (judge verdict: Accepted)

use std::cmp::Ordering;
use std::collections::BinaryHeap;
use std::collections::BTreeSet;
use std::cmp::Reverse;
use proconio::input;
use proconio::marker::Usize1;

fn main() {
    input! {
        n: usize,
        m: usize,
        s: Usize1,
        t: Usize1,
        mat: [(Usize1,Usize1,usize,usize); m]
    }

    let mut yen = vec![vec![]; n];
    let mut snu = vec![vec![]; n];
    for &(u,v,a,b) in mat.iter() {
        yen[u].push((v,a));
        yen[v].push((u,a));
        snu[u].push((v,b));
        snu[v].push((u,b));
    }

    let yeN = dijkstra(n, &yen, s);
    let snU = dijkstra(n, &snu, t);
    let mut ans = {
        let mut ans = Vec::new();
        for i in 0..n {
            ans.push((i,yeN[i]+snU[i]));
        }
        ans.sort_by(|(_,x),(_,y)| y.cmp(x));
        ans
    };

    let mut now = 0;
    while let Some((i,val)) = ans.pop() {
        if now <= i {
            for _ in now..=i {
                println!("{}",1000000000000000-val)
            }
            now = i+1;
        }
    }

}

#[derive(Clone,Copy)]
struct Pair {
    label: usize,
    value: usize,
}

impl Pair {
    fn new(l: usize, v: usize) -> Pair {
        Pair { label: l, value: v }
    } 
}

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

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

impl PartialEq for Pair {
    fn eq(&self, other: &Self) -> bool {
        (self.label,self.value) == (other.label,other.value)
    }
}

impl Eq for Pair { }

pub fn dijkstra (size: usize, list: &Vec<Vec<(usize,usize)>>, start: usize) -> Vec<usize> {
    let mut potential = vec![std::usize::MAX; size];
    potential[start] = 0;
    let mut heap = BinaryHeap::new();
    heap.push(Reverse(Pair::new(start,0)));
    let mut set = BTreeSet::new();
    for i in 0..size {
        set.insert(i);
    }
    while let Some(Reverse(pair)) = heap.pop() {
        if set.contains(&pair.label) {
            set.remove(&pair.label);
            for &(v,weight) in list[pair.label].iter() {
                if set.contains(&v) && (potential[v] > potential[pair.label] + weight) {
                    potential[v] = potential[pair.label] + weight;
                    heap.push(Reverse(Pair::new(v,potential[v])));
                }
            }
        }
        if set.is_empty() {
            break;
        }
    }
    potential
}