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++14 (GCC 5.4.1) (judge verdict: Accepted)

#define _USE_MATH_DEFINES
#include <bits/stdc++.h>
#include <vector>
#include <iostream>
#include<algorithm>
#include<string>
#include <map>
#include <queue>
#include <stack>
#include<set>
#include<math.h>

#define DIV 1000000007
using namespace std;
using ll = long long;

int main(void) {
	int n, m, s, t;
	cin >> n >> m >> s >> t;

	vector<vector<pair<int, pair<ll, ll>>>> city(n + 1);
	for (int i = 0; i < m; i++) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		city[u].push_back(make_pair(v, make_pair(a, b)));
		city[v].push_back(make_pair(u, make_pair(a, b)));
	}

	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
	vector<ll> yen(n + 1, 1e15);
	vector<bool> visit(n + 1);
	pq.emplace(0, s);
	yen[s] = 0;

	while (!pq.empty()) {
		pair<ll, int> x = pq.top();
		visit[x.second] = true;
		pq.pop();

		for (auto edge : city[x.second]) {
			if (visit[edge.first]) continue;

			if (yen[edge.first] > yen[x.second] + edge.second.first) {
				yen[edge.first] = yen[x.second] + edge.second.first;
				pq.emplace(yen[edge.first], edge.first);
			}

		}

	}
	
	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq2;
	vector<ll> su(n + 1, 1e15);
	vector<bool> visit2(n + 1);
	pq2.emplace(0, t);
	su[t] = 0;

	while (!pq2.empty()) {
		pair<ll, int> x = pq2.top();
		visit2[x.second] = true;
		pq2.pop();

		for (auto edge : city[x.second]) {
			if (visit2[edge.first]) continue;

			if (su[edge.first] > su[x.second] + edge.second.second) {
				su[edge.first] = su[x.second] + edge.second.second;
				pq2.emplace(su[edge.first], edge.first);
			}

		}

	}
	/*
	for (int i = 1; i <= n; i++) {
		cout << "i=" << i << " ";
		cout << yen[i] << " ";
		cout << su[i] << endl;
	}
	*/

	priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq3;
	for (int i = 1; i <= n; i++) {
		pq3.emplace(yen[i] + su[i], i);
	}

	for (int i = 0; i < n; i++) {
		while (pq3.top().second <= i) {
			pq3.pop();
		}
		cout << (ll)1e15 - pq3.top().first << endl;
	}
}