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 Java8 (OpenJDK 1.8.0) (judge verdict: Time Limit Exceeded)

import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.PriorityQueue;
import java.util.Scanner;

/*
 * SoundHound Inc. Programming Contest 2018 -Masters Tournament- D "Saving Snuuk"
 *   https://soundhound2018-summer-qual.contest.atcoder.jp/tasks/soundhound2018_summer_qual_d/
 */
public class Main {

	public static void main(String[] args) {

		InputScanner in = new InputScanner(System.in);
		PrintWriter out = new PrintWriter(System.out);

		int n = in.nextInt();
		int m = in.nextInt();
		int s = in.nextInt();
		int t = in.nextInt();

		Main main = new Main();
		Graph aGraph = main.new Graph();
		Graph bGraph = main.new Graph();
		for (int i = 1; i <= n; i++) {
			aGraph.addVertex(i);
			bGraph.addVertex(i);
		}
		for (int i = 1; i <= m; i++) {
			int u = in.nextInt();
			int v = in.nextInt();
			int a = in.nextInt();
			int b = in.nextInt();
			aGraph.addEdge(u, v, a);
			bGraph.addEdge(u, v, b);
		}

		for (int i = 0; i < n; i++) {

			long min = Long.MAX_VALUE;
			for (int j = i; j < n; j++) {
				long aCost = dijkstra(aGraph, s, j + 1);
				long bCost = dijkstra(bGraph, j + 1, t);
				min = Math.min(min, aCost + bCost);
			}

			long ans = (long) Math.pow(10, 15) - min;
			out.println(ans);
		}

		in.close();
		out.close();
	}

	static long dijkstra(Graph graph, int s, int t) {

		long[] cost = new long[graph.vertexNum];
		Arrays.fill(cost, Long.MAX_VALUE);
		cost[s - 1] = 0;

		PriorityQueue<long[]> pq = new PriorityQueue<long[]>(new Comparator<long[]>() {
			public int compare(long[] a, long[] b) {
				return Long.signum(a[1] - b[1]);
			}
		});
		pq.offer(new long[]{s, 0});

		while (!pq.isEmpty()) {

			long[] city = pq.poll();
			int currentNo = (int) city[0];
			long currentCost = city[1];

			if (currentNo == t) {
				break;
			}

			if (cost[currentNo - 1] < currentCost) {
				continue;
			}

			Vertex cityV = graph.map.get(currentNo);
			for (Edge edge : cityV.edges) {

				Vertex nextV = edge.vertex;
				int moveCost = edge.weight;

				if (cost[nextV.id - 1] > currentCost + moveCost) {
					cost[nextV.id - 1] = currentCost + moveCost;
					pq.offer(new long[]{nextV.id, cost[nextV.id - 1]});
				}

			}

		}

		return cost[t - 1];
	}

	class Edge {

		int weight;
		Vertex vertex;

		public Edge(Vertex vertex, int weight) {
			this.vertex = vertex;
			this.weight = weight;
		}
	}

	class Vertex {

		int id;
		List<Edge> edges;

		public Vertex(int id) {
			this.id = id;
			this.edges = new ArrayList<Edge>();
		}

		public void addEdge(Vertex vertex, int weight) {
			edges.add(new Edge(vertex, weight));
		}

	}

	class Graph {

		int vertexNum;
		int edgeNum;
		Map<Integer, Vertex> map;

		public Graph() {
			this.vertexNum = 0;
			this.edgeNum = 0;
			this.map = new HashMap<Integer, Vertex>();
		}

		public void addVertex(int vertex) {

			if (!map.containsKey(vertex)) {
				map.put(vertex, new Vertex(vertex));
				vertexNum++;
			}

		}

		public void addEdge(int fromVertex, int toVertex, int weight) {

			if (!map.containsKey(fromVertex)) {
				map.put(fromVertex, new Vertex(fromVertex));
				vertexNum++;
			}
			if (!map.containsKey(toVertex)) {
				map.put(toVertex, new Vertex(toVertex));
				vertexNum++;
			}

			Vertex fromV = map.get(fromVertex);
			Vertex toV = map.get(toVertex);
			fromV.addEdge(toV, weight);
			toV.addEdge(fromV, weight);
			edgeNum += 2;

		}

	}

	/*
	 * extends java.util.Scanner
	 */
	static class InputScanner {

		Scanner scanner;

		public InputScanner(InputStream inputStream) {
			this.scanner = new Scanner(inputStream);
		}

		public int nextInt() {
			return scanner.nextInt();
		}

		public long nextLong() {
			return scanner.nextLong();
		}

		public double nextDouble() {
			return scanner.nextDouble();
		}
		public String next() {
			return scanner.next();
		}

		public int[] nextIntArray(int n) {
			int[] ret = new int[n];
			for (int i = 0; i < n; i++) {
				ret[i] = scanner.nextInt();
			}
			return ret;
		}

		public int[][] nextIntArray(int h, int w) {
			int[][] ret = new int[h][w];
			for (int i = 0; i < h; i++) {
				for (int j = 0; j < w; j++) {
					ret[i][j] = scanner.nextInt();
				}
			}
			return ret;
		}

		public long[] nextLongArray(int n) {
			long[] ret = new long[n];
			for (int i = 0; i < n; i++) {
				ret[i] = scanner.nextLong();
			}
			return ret;
		}

		public long[][] nextlongArray(int h, int w) {
			long[][] ret = new long[h][w];
			for (int i = 0; i < h; i++) {
				for (int j = 0; j < w; j++) {
					ret[i][j] = scanner.nextLong();
				}
			}
			return ret;
		}

		public double[] nextDoubleArray(int n) {
			double[] ret = new double[n];
			for (int i = 0; i < n; i++) {
				ret[i] = scanner.nextDouble();
			}
			return ret;
		}

		public double[][] nextDoubleArray(int h, int w) {
			double[][] ret = new double[h][w];
			for (int i = 0; i < h; i++) {
				for (int j = 0; j < w; j++) {
					ret[i][j] = scanner.nextLong();
				}
			}
			return ret;
		}

		public String[] nextStringArray(int n) {
			String[] ret = new String[n];
			for (int i = 0; i < n; i++) {
				ret[i] = scanner.next();
			}
			return ret;
		}

		public String[] nextStringArray(String split) {
			String[] ret = scanner.next().split(split);
			return ret;
		}

		public String[][] nextStringArray(int h, int w) {
			String[][] ret = new String[h][w];
			for (int i = 0; i < h; i++) {
				for (int j = 0; j < w; j++) {
					ret[i][j] = scanner.next();
				}
			}
			return ret;
		}

		public String[][] nextStringArray(int h, String split) {
			String[][] ret = new String[h][];
			for (int i = 0; i < h; i++) {
				ret[i] = scanner.next().split(split);
			}
			return ret;
		}

		public void close() {
			scanner.close();
		}

	}

}