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)

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

#[derive(Debug, Copy, Clone)]
struct Edge {
    to: usize,
    yen: usize,
    snook: usize,
}

fn shortest_path_yen(adjl: &Vec<Vec<Edge>>, start: usize) -> Vec<usize> {
    let mut dist: Vec<_> = (0..adjl.len()).map(|_| std::usize::MAX).collect();

    let mut heap = std::collections::BinaryHeap::new();

    dist[start] = 0;
    heap.push(State {
        cost: 0,
        position: start,
    });

    while let Some(State { cost, position }) = heap.pop() {
        if cost > dist[position] {
            continue;
        }

        for edge in &adjl[position] {
            let next = State {
                cost: cost + edge.yen,
                position: edge.to,
            };
            if next.cost < dist[next.position] {
                heap.push(next);
                dist[next.position] = next.cost;
            }
        }
    }
    dist
}
fn shortest_path_snook(adjl: &Vec<Vec<Edge>>, start: usize) -> Vec<usize> {
    let mut dist: Vec<_> = (0..adjl.len()).map(|_| std::usize::MAX).collect();

    let mut heap = std::collections::BinaryHeap::new();

    dist[start] = 0;
    heap.push(State {
        cost: 0,
        position: start,
    });

    while let Some(State { cost, position }) = heap.pop() {
        if cost > dist[position] {
            continue;
        }

        for edge in &adjl[position] {
            let next = State {
                cost: cost + edge.snook,
                position: edge.to,
            };
            if next.cost < dist[next.position] {
                heap.push(next);
                dist[next.position] = next.cost;
            }
        }
    }
    dist
}

fn main() {
    let (v, e, start, terminal): (usize, usize, usize, usize) =
        (read(), read(), read::<usize>() - 1, read::<usize>() - 1);
    let mut adjl: Vec<Vec<Edge>> = vec![vec![]; v];
    for _ in 0..e {
        let (u, v, yen, snook): (usize, usize, usize, usize) =
            (read::<usize>() - 1, read::<usize>() - 1, read(), read());
        adjl[u].push(Edge {
            to: v,
            yen: yen,
            snook: snook,
        });
        adjl[v].push(Edge {
            to: u,
            yen: yen,
            snook: snook,
        });
    }
    let yd = shortest_path_yen(&adjl, start);
    let sd = shortest_path_snook(&adjl, terminal);
    let mut tmpcost = std::usize::MAX;
    let mut costs = vec![];
    for i in (0..v).rev() {
        tmpcost = min(tmpcost, yd[i] + sd[i]);
        costs.push(tmpcost);
    }
    let m = 10usize.pow(15);
    for i in 0..v {
        println!("{}", m - costs[v - i - 1]);
    }
}

// =========
#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{HashMap, HashSet};
#[allow(unused_imports)]
use std::process::exit;

#[allow(dead_code)]
const MOD: usize = 1000000007;

fn read<T: std::str::FromStr>() -> T {
    use std::io::Read;
    let stdin = std::io::stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

// =========