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 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>
int u[200005], v[200005];
long long int a[200005], b[200005];
int asc(const void* x, const void* y)
{
	if (u[*(int*)x] > u[*(int*)y])return 1;
	if (u[*(int*)x] < u[*(int*)y])return -1;
	return 0;
}
int h[1000006], l;
long long int dist[100005];
void swap_h(int x, int y)
{
	int f = h[x];
	h[x] = h[y];
	h[y] = f;
	return;
}
int comp_h(int x, int y)
{
	x = h[x];
	y = h[y];
	if (dist[x] > dist[y])return 1;
	if (dist[x] < dist[y])return -1;
	return 0;
}
void f1(int n)
{
	h[l] = n;
	int now = l;
	l++;
	while (now > 0)
	{
		if (comp_h((now - 1) / 2, now) > 0)
			swap_h((now - 1) / 2, now);
		now = (now - 1) / 2;
	}
	return;
}
int f2()
{
	l--;
	int now = 0;
	swap_h(0, l);
	for (;;)
	{
		if (2 * now + 2 < l)
		{
			if (comp_h(2 * now + 1, 2 * now + 2) > 0)
			{
				if (comp_h(now, 2 * now + 2) > 0)
					swap_h(now, 2 * now + 2);
				now = 2 * now + 2;
			}
			else
			{
				if (comp_h(now, 2 * now + 1) > 0)
					swap_h(now, 2 * now + 1);
				now = 2 * now + 1;
			}
		}
		else if (2 * now + 1 < l)
		{
			if (comp_h(now, 2 * now + 1) > 0)
				swap_h(now, 2 * now + 1);
			now = 2 * now + 1;
		}
		else
			break;
	}
	return h[l];
}
int main()
{
	int n, m, s, t;
	scanf("%d %d %d %d", &n, &m, &s, &t);
	s--;
	t--;
	int i;
	for (i = 0; i < m; i++)
	{
		scanf("%d %d %lld %lld", &u[2 * i], &v[2 * i], &a[2 * i], &b[2 * i]);
		u[2 * i]--;
		v[2 * i]--;
		u[2 * i + 1] = v[2 * i];
		v[2 * i + 1] = u[2 * i];
		a[2 * i + 1] = a[2 * i];
		b[2 * i + 1] = b[2 * i];
	}
	int c[200005];
	for (i = 0; i < 2 * m; i++)
		c[i] = i;
	qsort(c, 2 * m, sizeof(int), asc);
	long long int ans[100005];
	for (i = 0; i < n; i++)
		ans[i] = 1000000000000000;
	int min, mid, max;
	for (i = 0; i < n; i++)
		dist[i] = 1000000000000000000;
	l = 0;
	dist[s] = 0;
	f1(s);
	while (l > 0)
	{
		i = f2();
		min = 0;
		max = 2 * m;
		while (max - min > 1)
		{
			mid = (max + min) / 2;
			if (u[c[mid]] < i)
				min = mid;
			else
				max = mid;
		}
		if (u[c[min]] < i)
			min++;
		if (min < 2 * m)
		{
			while (u[c[min]] == i)
			{
				if (dist[v[c[min]]] > dist[i] + a[c[min]])
				{
					dist[v[c[min]]] = dist[i] + a[c[min]];
					f1(v[c[min]]);
				}
				min++;
				if (min == 2 * m)
					break;
			}
		}
	}
	for (i = 0; i < n; i++)
		ans[i] -= dist[i];
	for (i = 0; i < n; i++)
		dist[i] = 1000000000000000000;
	l = 0;
	dist[t] = 0;
	f1(t);
	while (l > 0)
	{
		i = f2();
		min = 0;
		max = 2 * m;
		while (max - min > 1)
		{
			mid = (max + min) / 2;
			if (u[c[mid]] < i)
				min = mid;
			else
				max = mid;
		}
		if (u[c[min]] < i)
			min++;
		if (min < 2 * m)
		{
			while (u[c[min]] == i)
			{
				if (dist[v[c[min]]] > dist[i] + b[c[min]])
				{
					dist[v[c[min]]] = dist[i] + b[c[min]];
					f1(v[c[min]]);
				}
				min++;
				if (min == 2 * m)
					break;
			}
		}
	}
	for (i = 0; i < n; i++)
		ans[i] -= dist[i];
	for (i = n - 2; i >= 0; i--)
		if (ans[i] < ans[i + 1])
			ans[i] = ans[i + 1];
	for (i = 0; i < n; i++)
		printf("%lld\n", ans[i]);
	return 0;
}