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 C (GCC 9.2.1) (judge verdict: Wrong Answer)

#include <stdio.h>
#include <stdlib.h>

typedef struct List {
	struct List *next;
	int v, cost, tmp;
} list;

typedef struct {
	int key, id;
} data;
 
typedef struct {
	data *obj;
	int size;
} min_heap;
 
void push(data x, min_heap* h)
{
	int i = h->size;
	data tmp;
	h->obj[(h->size)++] = x;
	while (1) {
		if (i == 0) break;
		else if (h->obj[i].key < h->obj[(i-1)/2].key) {
			tmp = h->obj[(i-1)/2];
			h->obj[(i-1)/2] = h->obj[i];
			h->obj[i] = tmp;
			i = (i - 1) / 2;
		} else break;
	}
}
 
data pop(min_heap* h)
{
	int i = 0, j;
	data output = h->obj[0], tmp;
	h->obj[0] = h->obj[--(h->size)];
	while (1) {
		if (i * 2 + 1 >= h->size) break;
		else if (i * 2 + 2 >= h->size) j = i * 2 + 1;
		else if (h->obj[i*2+1].key < h->obj[i*2+2].key) j = i * 2 + 1;
		else j = i * 2 + 2;
		if (h->obj[j].key < h->obj[i].key) {
			tmp = h->obj[j];
			h->obj[j] = h->obj[i];
			h->obj[i] = tmp;
			i = j;
		} else break;
	}
	return output;
}

long long* dijkstra(list **adj, int s, int n, int m)
{
	int i, *prev = (int*)malloc(sizeof(int) * (n + 1));
	long long *dist = (long long*)malloc(sizeof(long long) * (n + 1));
	for (i = 1; i <= n; i++) {
		dist[i] = (long long)1 << 60;
		prev[i] = 0;
	}
	dist[s] = 0;
	prev[s] = -s;
	
	int u, w;
	list *p;
	data d;
	min_heap h;
	h.obj = (data*)malloc(sizeof(data) * (m * 2 + 1));
	h.size = 0;
	d.key = 0;
	d.id = s;
	push(d, &h);
	while (h.size > 0) {
		u = pop(&h).id;
		if (prev[u] > 0) continue;
		prev[u] *= -1;
		for (p = adj[u]; p != NULL; p = p->next) {
			w = p->v;
			if (dist[u] + p->cost < dist[w]) {
				dist[w] = dist[u] + p->cost;
				prev[w] = -u;
				d.key = dist[w];
				d.id = w;
				push(d, &h);
			}
		}
	}
	
	free(prev);
	free(h.obj);
	return dist;
}

int main()
{
	int i, n, m, s, t, u, w, a, b;
	scanf("%d %d %d %d", &n, &m, &s, &t);
	list **adj = (list**)malloc(sizeof(list*) * (n + 1)), *e = (list*)malloc(sizeof(list) * m * 2);
	for (i = 1; i <= n; i++) adj[i] = NULL;
	for (i = 0; i < m; i++) {
		scanf("%d %d %d %d", &u, &w, &a, &b);
		e[i*2].v = w;
		e[i*2+1].v = u;
		e[i*2].cost = a;
		e[i*2+1].cost = a;
		e[i*2].tmp = b;
		e[i*2+1].tmp = b;
		e[i*2].next = adj[u];
		e[i*2+1].next = adj[w];
		adj[u] = &(e[i*2]);
		adj[w] = &(e[i*2+1]);
	}
	
	long long *dist[2];
	dist[0] = dijkstra(adj, s, n, m);
	for (i = 0; i < m * 2; i++) e[i].cost = e[i].tmp;
	dist[1] = dijkstra(adj, t, n, m);
	
	data d;
	min_heap h;
	h.size = 0;
	h.obj = (data*)malloc(sizeof(data) * (n + 1));
	for (i = 1; i <= n; i++) {
		d.key = dist[0][i] + dist[1][i];
		d.id = i;
		push(d, &h);
	}
	
	d.id = 0;
	for (i = 0; i < n; i++) {
		while (d.id <= i) d = pop(&h);
		printf("%lld\n", 1000000000000000 - d.key);
	}
	fflush(stdout);
	return 0;
}