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.

Examples

Input

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1


Output

999999999999998
999999999999989
999999999999979
999999999999897


Input

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


Output

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
import java.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Arrays;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.ArrayList;
import java.util.AbstractCollection;
import java.util.List;
import java.util.Comparator;

/**
 * Built using CHelper plug-in Actual solution is at the top
 *
 * @author Silviase
 */
public class Main {

  public static void main(String[] args) {
    InputStream inputStream = System.in;
    OutputStream outputStream = System.out;
    Scanner in = new Scanner(inputStream);
    PrintWriter out = new PrintWriter(outputStream);
    DSavingSnuuk solver = new DSavingSnuuk();
    solver.solve(1, in, out);
    out.close();
  }

  static class DSavingSnuuk {

    public void solve(int testNumber, Scanner in, PrintWriter out) {
      int n = in.nextInt();
      int m = in.nextInt();
      int from = in.nextInt() - 1;
      int to = in.nextInt() - 1;
      Graph yen = new Graph(n);
      Graph snu = new Graph(n);

      for (int i = 0; i < m; i++) {
        int f = in.nextInt() - 1;
        int t = in.nextInt() - 1;
        int cy = in.nextInt();
        int cs = in.nextInt();
        yen.addUndirectedEdge(f, t, cy);
        snu.addUndirectedEdge(f, t, cs);
      }

      long[] costFromStartByYen = yen.dijkstra(from);
      long[] costToGoalBySnuuk = snu.dijkstra(to);

      PriorityQueue<Pair<Long, Integer>> totalCost =
          new PriorityQueue<>(Comparator.comparing(x -> x.getX()));
      for (int i = 0; i < n; i++) {
        totalCost.add(new Pair<>(costFromStartByYen[i] + costToGoalBySnuuk[i], i));
      }

      for (int i = 0; i < n; i++) {
        while (!totalCost.isEmpty() && totalCost.peek().getY() < i) {
          totalCost.poll();
        }
        if (!totalCost.isEmpty()) {
          out.println((long) 1e15 - totalCost.peek().getX());
        }
      }

    }

  }

  static class Pair<T, U> {

    private T x;
    private U y;

    public Pair(T t, U u) {
      this.x = t;
      this.y = u;
    }

    public T getX() {
      return x;
    }

    public U getY() {
      return y;
    }

    public String toString() {
      return this.x.toString() + " " + this.y.toString();
    }

    public int hashCode() {
      return super.hashCode();
    }

  }

  static class Edge {

    int to;
    int cost = 1;

    public Edge(int to) {
      this.to = to;
    }

    public Edge(int to, int cost) {
      this.to = to;
      this.cost = cost;
    }

    public String toString() {
      return "to :" + to + ", cost :" + cost;
    }

  }

  static class Vertex {

    private List<Edge> adj;

    public Vertex() {
      this.adj = new ArrayList<>();
    }

    public void addEdge(Edge e) {
      adj.add(e);
    }

    public List<Edge> getAdj() {
      return adj;
    }

    public int getSize() {
      return adj.size();
    }

  }

  static class Graph {

    private Vertex[] v;
    private int size;

    public Graph(int size) {
      this.size = size;
      v = new Vertex[this.size];
      for (int i = 0; i < this.size; i++) {
        v[i] = new Vertex();
      }
    }

    public void addUndirectedEdge(int from, int to, int cost) {
      v[from].addEdge(new Edge(to, cost));
      v[to].addEdge(new Edge(from, cost));
    }

    public long[] dijkstra(int from) {
      PriorityQueue<Pair<Long, Integer>> pq =
          new PriorityQueue<>(Comparator.comparing(x -> x.getX()));
      long[] d = new long[size];
      Arrays.fill(d, (long) 1e20);
      d[from] = 0;
      pq.add(new Pair<>((long) 0, from));
      while (!pq.isEmpty()) {
        Pair ip = pq.peek();
        pq.poll();
        long c = (long) ip.getX();
        int now = (int) ip.getY();
        if (d[now] < c) {
          continue;
        }
        for (int i = 0; i < v[now].getSize(); i++) {
          Edge e = v[now].getAdj().get(i);
          // System.out.println(e.toString());
          if (d[e.to] > d[now] + e.cost) {
            d[e.to] = d[now] + e.cost;
            Pair<Long, Integer> p = new Pair<>(d[e.to], e.to);
            pq.add(p);
            // System.out.println("d[" + e.to + "] update");
          }
        }
      }
      return d;
    }

  }
}

