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: Wrong Answer)

#![allow(unused_parens)]
#![allow(unused_imports)]
#![allow(non_upper_case_globals)]
#![allow(non_snake_case)]
#![allow(unused_mut)]
#![allow(unused_variables)]
#![allow(dead_code)]

use itertools::Itertools;
use proconio::input;
use proconio::marker::{Chars, Usize1};

#[allow(unused_macros)]
#[cfg(debug_assertions)]
macro_rules! mydbg {
    //($arg:expr) => (dbg!($arg))
    //($arg:expr) => (println!("{:?}",$arg));
      ($($a:expr),*) => {
          eprintln!(concat!($(stringify!($a), " = {:?}, "),*), $($a),*);
      }
}
#[cfg(not(debug_assertions))]
macro_rules! mydbg {
    ($($arg:expr),*) => {};
}

macro_rules! echo {
    ($($a:expr),*) => {
          $(println!("{}",$a))*
      }
}

use std::cmp::*;
use std::collections::*;
use std::ops::{Add, Div, Mul, Sub};

#[allow(dead_code)]
static INF_I64: i64 = 92233720368547758;
#[allow(dead_code)]
static INF_I32: i32 = 21474836;
#[allow(dead_code)]
static INF_USIZE: usize = 18446744073709551;
#[allow(dead_code)]
static M_O_D: usize = 1000000007;
#[allow(dead_code)]
static PAI: f64 = 3.1415926535897932;

trait IteratorExt: Iterator {
    fn toVec(self) -> Vec<Self::Item>;
}

impl<T: Iterator> IteratorExt for T {
    fn toVec(self) -> Vec<Self::Item> {
        self.collect()
    }
}

trait CharExt {
    fn toNum(&self) -> usize;
    fn toAlphabetIndex(&self) -> usize;
    fn toNumIndex(&self) -> usize;
}
impl CharExt for char {
    fn toNum(&self) -> usize {
        return *self as usize;
    }
    fn toAlphabetIndex(&self) -> usize {
        return self.toNum() - 'a' as usize;
    }
    fn toNumIndex(&self) -> usize {
        return self.toNum() - '0' as usize;
    }
}

trait VectorExt {
    fn joinToString(&self, s: &str) -> String;
}
impl<T: ToString> VectorExt for Vec<T> {
    fn joinToString(&self, s: &str) -> String {
        return self
            .iter()
            .map(|x| x.to_string())
            .collect::<Vec<_>>()
            .join(s);
    }
}

trait StringExt {
    fn get_reverse(&self) -> String;
}
impl StringExt for String {
    fn get_reverse(&self) -> String {
        self.chars().rev().collect::<String>()
    }
}

trait UsizeExt {
    fn pow(&self, n: usize) -> usize;
}
impl UsizeExt for usize {
    fn pow(&self, n: usize) -> usize {
        return ((*self as u64).pow(n as u32)) as usize;
    }
}

pub struct Dikstra {
    dist: Vec<i64>,
    start: usize,
}
impl Dikstra {
    pub fn new(g: &Vec<Vec<(i64, usize)>>, start: usize) -> Self {
        let mut d = Dikstra {
            dist: vec![INF_I64; g.len()],
            start: start,
        };
        d.calc(g);
        return d;
    }
    fn calc(&mut self, g: &Vec<Vec<(i64, usize)>>) {
        let mut used = vec![false; g.len()];
        let mut kouho = std::collections::BinaryHeap::new();
        kouho.push((0, self.start));
        while !kouho.is_empty() {
            let (_, k) = kouho.pop().unwrap();
            used[self.start] = true;
            self.dist[self.start] = 0;
            for i in 0..g[k].len() {
                let (d, p) = g[k][i];
                if used[p] {
                    continue;
                }
                if self.dist[p] > self.dist[k] + d as i64 {
                    self.dist[p] = self.dist[k] + d as i64;
                    kouho.push((self.dist[p] * -1, p));
                }
            }
        }
    }
    pub fn dist_to(&mut self, x: usize) -> i64 {
        return self.dist[x];
    }
}

macro_rules ! max { ( $ x : expr ) => ( $ x ) ; ( $ x : expr , $ ( $ xs : expr ) ,+ ) => { { use std :: cmp :: max ; max ( $ x , max ! ( $ ( $ xs ) ,+ ) ) } } ; }
macro_rules ! min { ( $ x : expr ) => ( $ x ) ; ( $ x : expr , $ ( $ xs : expr ) ,+ ) => { { use std :: cmp :: min ; min ( $ x , min ! ( $ ( $ xs ) ,+ ) ) } } ; }
fn main() {
    input! {
       N: usize,
       M:usize,
       S:Usize1,
       T:Usize1,
    }

    let mut mapS = vec![vec![]; N];
    let mut mapG = vec![vec![]; N];
    for _ in 0..M {
        input! {
            u:Usize1,
            v:Usize1,
            a:i64,
            b:i64,
        }
        mapS[u].push((a, v));
        mapS[v].push((a, u));
        mapG[u].push((b, v));
        mapG[v].push((b, u));
    }

    let mut ds = Dikstra::new(&mapS, S);
    let mut dt = Dikstra::new(&mapG, T);
    let mut tmp = vec![0; N];
    for i in 0..N {
        tmp[i] = ds.dist_to(i) + dt.dist_to(i);
    }
    mydbg!(tmp);
    let mut ans = vec![INF_I64; N];
    ans[N - 1] = 10_i64.pow(15) - tmp[N - 1];
    for i in (0..N - 1).rev() {
        ans[i] = 10_i64.pow(15) - min!(tmp[i], ans[i + 1], tmp[T]);
    }
    mydbg!(ans);

    echo!(ans.joinToString("\n"));
}