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: Time Limit Exceeded)

fn main() {
    let mut nmst = String::new();
    std::io::stdin().read_line(&mut nmst).unwrap();
    let nmst: Vec<usize> = nmst.split_whitespace().map(|x|x.parse().unwrap()).collect();
    let n = nmst[0];
    let m = nmst[1];
    let s = nmst[2]-1;
    let t = nmst[3]-1;
    let mut g: Vec<Vec<(usize,usize,usize)>> = vec![vec![];n];
    for _ in 0..m {
        let mut uvab = String::new();
        std::io::stdin().read_line(&mut uvab).unwrap();
        let uvab: Vec<usize> = uvab.split_whitespace().map(|x|x.parse().unwrap()).collect();
        let u = uvab[0]-1;
        let v = uvab[1]-1;
        let a = uvab[2];
        let b = uvab[3];
        g[u].push((v,a,b));
        g[v].push((u,a,b));
    }
    let mut d = vec![0usize;n];
    let mut e = vec![(0usize,n);n];
    let mut q = std::collections::BinaryHeap::new();
    q.reserve(2*n);
    q.push((1e15 as usize,n,s));
    let mut r = vec![]; r.reserve(n);
    while let Some((y,c,v)) = q.pop() {
        if c==n {
            if d[v]>=y {continue;}
            d[v]=y;
            if e[v].0<y || e[v].1<v {
                q.push((y,v,v));
            }
            for &(u,a,_) in &g[v] { if d[u]<y-a {
                q.push((y-a,c,u));
            }}
        }
        else {
            if e[v].0>=y && e[v].1>=c {continue;}
            if v==t {
                r.resize(c+1,y);
                if r.len()>=n {break;}
            }
            e[v].0=y;
            e[v].1=c;
            for &(u,_,b) in &g[v] { if e[u].0<y-b || e[u].1<c {
                q.push((y-b,c,u));
            }}
        }
    }
    for y in r {
        println!("{}",y);
    }
}