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: Accepted)

#define _USE_MATH_DEFINES

#include <iostream>
#include <iomanip>
#include <string>
#include <vector>
#include <algorithm>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <cmath>
//#include <atcoder/all>
using namespace std;
//using namespace atcoder;

typedef long long ll;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<ll, ll, ll> tl3;

//typedef modint998244353 mint;

const int BIG_NUM = 1e9;
const ll INF = 1000000000000000000;
const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;

const ll MAX = 1e9 + 5;


int main() {
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	s--;
	t--;
	vector<vector<pll>> g1(n);
	vector<vector<pll>> g2(n);
	for (int i = 0; i < m; i++) {
		ll u, v, a, b;
		cin >> u >> v >> a >> b;
		u--;
		v--;
		g1[u].emplace_back(make_pair(v, a));
		g1[v].emplace_back(make_pair(u, a));
		g2[u].emplace_back(make_pair(v, b));
		g2[v].emplace_back(make_pair(u, b));
	}

	vector<ll> d1(n, INF);
	vector<ll> d2(n, INF);

	d1[s] = 0;
	d2[t] = 0;
	priority_queue<pll, vector<pll>, greater<pll>> pq;
	vector<bool> ic1(n);
	pq.push(make_pair(0, s));
	while (!pq.empty()) {
		int v = pq.top().second;
		pq.pop();
		if (ic1[v]) {
			continue;
		}
		
		for (pii e : g1[v]) {
			int u = e.first;
			int c = e.second;

			if (d1[u] > d1[v] + c) {
				d1[u] = d1[v] + c;
				pq.push(make_pair(d1[u], u));
			}
		}

		ic1[v] = true;
	}

	vector<bool> ic2(n);
	pq.push(make_pair(0, t));
	while (!pq.empty()) {
		int v = pq.top().second;
		pq.pop();
		if (ic2[v]) {
			continue;
		}

		for (pii e : g2[v]) {
			int u = e.first;
			int c = e.second;

			if (d2[u] > d2[v] + c) {
				d2[u] = d2[v] + c;
				pq.push(make_pair(d2[u], u));
			}
		}

		ic2[v] = true;
	}

	vector<ll> ans(n);
	ll a = 0;
	for (int i = n - 1; i >= 0; i--) {

		a = max(a,(ll) 1e15 - (d1[i] + d2[i]));
		ans[i] = a;
	}

	for (ll ai : ans) {
		cout << ai << endl;
	}
}