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)

#include "iostream"
#include "climits"
#include "list"
#include "queue"
#include "stack"
#include "set"
#include "functional"
#include "algorithm"
#include "string"
#include "map"
#include "unordered_map"
#include "unordered_set"
#include "iomanip"
#include "cmath"
#include "random"
#include "bitset"
#include "cstdio"

using namespace std;

const long long int MOD = 1000000007;
const long double EPS = 1e-9;

long long int N, M, K, H, W, L, R;

struct Edge {
	long long int to;
	long long int yen;
	long long int snuk;
};

int main() {
	ios::sync_with_stdio(false);
	cin.tie(0);

	cin >> N >> M >> L >> R;
	vector<long long int>s(N + 1, MOD*MOD);
	vector<long long int>y(N + 1, MOD*MOD);
	vector<vector<Edge>>edge(N + 1);
	for (int i = 0; i < M; i++) {
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		edge[a].push_back({ b,c,d });
		edge[b].push_back({ a,c,d });
	}
	s[R] = 0;
	set<pair<long long int, int>>st;
	st.insert({ 0,R });
	while (!st.empty()) {
		auto c = *st.begin();
		st.erase(st.begin());
		if (s[c.second] < c.first)continue;
		for (auto i : edge[c.second]) {
			if (s[i.to] > s[c.second] + i.snuk) {
				s[i.to] = s[c.second] + i.snuk;
				st.insert({ s[i.to],i.to });
			}
		}
	}
	y[L] = 0;
	st.insert({ 0,L });
	while (!st.empty()) {
		auto c = *st.begin();
		st.erase(st.begin());
		if (y[c.second] < c.first)continue;
		for (auto i : edge[c.second]) {
			if (y[i.to] > y[c.second] + i.yen) {
				y[i.to] = y[c.second] + i.yen;
				st.insert({ y[i.to],i.to });
			}
		}
	}
	set<pair<long long int, int>>ans;
	for (int i = 1; i <= N; i++) {
		ans.insert({ y[i] + s[i],i });
	}
	for (int i = 1; i <= N; i++) {
		while ((*(ans.begin())).second < i) {
			ans.erase(ans.begin());
		}
		cout << 1000000000000000 - (*ans.begin()).first << endl;
	}
	return 0;
}