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)

#[derive(Clone, Copy, PartialEq, Eq)]
struct Pointer(usize);

impl Pointer {
    const NULL: usize = std::usize::MAX;
    fn new(p: usize) -> Self {
        Pointer(p)
    }
    fn set(&mut self, x: usize) {
        self.0 = x;
    }
    fn get(&self) -> Option<usize> {
        if self.is_null() {
            None
        } else {
            Some(self.0)
        }
    }
    fn clear(&mut self) {
        self.set(Self::NULL);
    }
    fn is_null(&self) -> bool {
        self.0 == Self::NULL
    }
}

impl Default for Pointer {
    fn default() -> Self {
        Pointer::new(Pointer::NULL)
    }
}

struct Node<K, V> {
    key: K,
    value: V,
    parent: Pointer,
    child: Pointer,
    left: Pointer,
    right: Pointer,
}

impl<K, V> Node<K, V> {
    fn new(key: K, value: V) -> Self {
        Node {
            key: key,
            value: value,
            parent: Pointer::default(),
            child: Pointer::default(),
            left: Pointer::default(),
            right: Pointer::default(),
        }
    }
    fn rooting(&mut self, x: usize) {
        self.parent.clear();
        self.left.set(x);
        self.right.set(x);
    }
}

struct Heap<K, V> {
    array: Vec<Node<K, V>>,
    free: Pointer,
    root: Pointer,
}

impl<K: Ord + Copy, V: Copy> Heap<K, V> {
    fn new() -> Self {
        Heap {
            array: vec![],
            free: Pointer::default(),
            root: Pointer::default(),
        }
    }
    fn meld(&mut self, x: Pointer, y: Pointer) -> Pointer {
        if x.is_null() {
            return y;
        }
        if y.is_null() {
            return x;
        }
        let mut x = x.get().unwrap();
        let mut y = y.get().unwrap();
        let node = &mut self.array;
        if node[x].key >= node[y].key {
            std::mem::swap(&mut x, &mut y);
        }
        node[y].parent.set(x);
        let child = node[x].child;
        node[x].child.set(y);
        if let Some(c) = child.get() {
            let r = node[c].right.get().unwrap();
            node[c].right.set(y);
            node[r].left.set(y);
            node[y].left.set(c);
            node[y].right.set(r);
        }
        Pointer::new(x)
    }
    fn merge_list(&mut self, lst: Pointer) -> Pointer {
        if lst.is_null() {
            return lst;
        }
        let k = lst.get().unwrap();
        let l = self.array[k].left.get().unwrap();
        self.array[l].right.clear();
        self.array[k].left.clear();
        let mut lst = lst;
        let mut root = Pointer::default();
        while let Some(a) = lst.get() {
            if let Some(b) = self.array[a].right.get() {
                lst = self.array[b].right;
                self.array[a].rooting(a);
                self.array[b].rooting(b);
                let x = self.meld(Pointer::new(a), Pointer::new(b));
                root = self.meld(root, x);
            } else {
                lst.clear();
                self.array[a].rooting(a);
                root = self.meld(root, Pointer::new(a));
            }
        }
        root
    }
    fn push(&mut self, key: K, value: V) -> Pointer {
        if self.free.is_null() {
            let x = self.array.len();
            self.free.set(x);
            self.array.push(Node::new(key, value));
        }
        let res = self.free;
        let x = res.get().unwrap();
        self.free = self.array[x].child;
        self.array[x] = Node::new(key, value);
        self.array[x].rooting(x);
        self.root = self.meld(self.root, res);
        res
    }
    fn pop(&mut self) -> Option<(K, V)> {
        let root = self.root;
        root.get().map(|root| {
            let lst = self.array[root].child;
            self.array[root].child = self.free;
            self.free.set(root);
            self.root = self.merge_list(lst);
            (self.array[root].key, self.array[root].value)
        })
    }
}

use proconio::*;
use proconio::marker::*;

fn dijkstra(s: usize, g: &[Vec<(usize, u64)>]) -> Vec<u64> {
    let inf = 10u64.pow(15);
    let mut dp = vec![inf; g.len()];
    dp[s] = 0;
    let mut h = Heap::new();
    h.push(0, s);
    while let Some((d, v)) = h.pop() {
        if d > dp[v] {
            continue;
        }
        for &(u, c) in g[v].iter() {
            let d = d + c;
            if dp[u] > d {
                dp[u] = d;
                h.push(d, u);
            }
        }
    }
    dp
}

#[fastout]
fn run() {
    input! {
        n: usize,
        m: usize,
        s: Usize1,
        t: Usize1,
        e: [(Usize1, Usize1, u64, u64); m],
    }
    let mut g = vec![vec![]; n];
    for &(a, b, c, _) in e.iter() {
        g[a].push((b, c));
        g[b].push((a, c));
    }
    let dp_s = dijkstra(s, &g);
    g.iter_mut().for_each(|g| g.clear());
    for &(a, b, _, c) in e.iter() {
        g[a].push((b, c));
        g[b].push((a, c));
    }
    let dp_t = dijkstra(t, &g);
    let mut ans = vec![0; n];
    let ini = 10u64.pow(15);
    let mut val = ini;
    for (ans, (s, t)) in ans.iter_mut().zip(dp_s.iter().zip(dp_t.iter())).rev() {
        val = std::cmp::min(*s + *t, val);
        *ans = ini - val;
    }
    for a in ans {
        println!("{}", a);
    }
}

fn main() {
    run();
}