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)

fn main() {
	let mut s = String::new();
	use std::io::Read;
	std::io::stdin().read_to_string(&mut s).unwrap();
	let mut s = s.split_whitespace();
	let n: usize = s.next().unwrap().parse().unwrap();
	let m: usize = s.next().unwrap().parse().unwrap();
	let src: usize = s.next().unwrap().parse::<usize>().unwrap() - 1;
	let dst: usize = s.next().unwrap().parse::<usize>().unwrap() - 1;
	let mut yen = vec![vec![]; n];
	let mut snuuk = vec![vec![]; n];
	for _ in 0..m {
		let src: usize = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let dst: usize = s.next().unwrap().parse::<usize>().unwrap() - 1;
		let yc: i64 = s.next().unwrap().parse().unwrap();
		let sc: i64 = s.next().unwrap().parse().unwrap();
		yen[src].push(Edge { to: dst, cost: yc });
		yen[dst].push(Edge { to: src, cost: yc });
		snuuk[src].push(Edge { to: dst, cost: sc });
		snuuk[dst].push(Edge { to: src, cost: sc });
	}
	let yen = dijkstra(&yen, src);
	let snuuk = dijkstra(&snuuk, dst);
	let ans: Vec<i64> = yen.iter().rev().zip(snuuk.iter().rev()).scan(std::i64::MAX, |s, (yen, snuuk)|{
		*s = std::cmp::min(*s, yen + snuuk);
		Some(*s)
	}).collect();
	use std::io::Write;
	let out = std::io::stdout();
	let mut out = std::io::BufWriter::new(out.lock());
	for ans in ans.into_iter().rev() {
		writeln!(out, "{}", 1e15 as i64 - ans).unwrap();
	}
}

fn dijkstra(edges: &Vec<Vec<Edge>>, src: usize) -> Vec<i64> {
	let mut dist = vec![std::i64::MAX; edges.len()];
	dist[src] = 0;
	let mut q = std::collections::BinaryHeap::new();
	q.push((0, src));
	while let Some((cost, i)) = q.pop() {
		let cost = -cost;
		if dist[i] < cost { continue; }
		for e in edges[i].iter() {
			if dist[e.to] <= cost + e.cost { continue; }
			dist[e.to] = cost + e.cost;
			q.push((-dist[e.to], e.to));
		}
	}
	dist
}

#[derive(Clone, Copy)]
struct Edge {
	to: usize,
	cost: i64,
}