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.util.ArrayList;
import java.util.Arrays;
import java.util.Comparator;
import java.util.PriorityQueue;
import java.util.Scanner;

public class Main {
    public static void main(String args[]) {
        try (Scanner in = new Scanner(System.in)) {
            int n = in.nextInt();
            int m = in.nextInt();
            int s = in.nextInt() - 1;
            int t = in.nextInt() - 1;

            ArrayList<Edge>[] edges = new ArrayList[n];
            for (int i = 0; i < edges.length; i++) {
                edges[i] = new ArrayList<>();
            }
            for (int i = 0; i < m; i++) {
                int u = in.nextInt() - 1;
                int v = in.nextInt() - 1;
                int a = in.nextInt();
                int b = in.nextInt();
                edges[u].add(new Edge(v, a, b));
                edges[v].add(new Edge(u, a, b));
            }

            long[] cost = new long[n];
            Arrays.fill(cost, (long) 1e18);
            dijkstra(s, true, n, edges, cost);
            long[] costS = new long[n];
            Arrays.fill(costS, (long) 1e18);
            dijkstra(t, !true, n, edges, costS);

            long[] mins = new long[n];
            Arrays.fill(mins, (long) 1e18);
            for (int i = n - 1; i >= 0; i--) {
                mins[i] = Math.min(mins[i], cost[i] + costS[i]);
                if (i + 1 < n) {
                    mins[i] = Math.min(mins[i], mins[i + 1]);
                }
            }
            for (int i = 0; i < n; i++) {
                System.out.println(1000000000000000L - mins[i]);
            }

        }
    }

    private static void dijkstra(int startVertex, boolean b, int n, ArrayList<Edge>[] edges, long[] cost) {

        PriorityQueue<Vertex> pq = new PriorityQueue<>(new Comparator<Vertex>() {
            @Override
            public int compare(Vertex o1, Vertex o2) {
                if (o1.cost < o2.cost) {
                    return -1;
                }
                if (o1.cost > o2.cost) {
                    return 1;
                }
                return 0;
            }
        });

        {
            pq.add(new Vertex(startVertex, 0));
            cost[startVertex] = 0;
        }
        for (; !pq.isEmpty();) {
            Vertex vertex = pq.poll();

            // if (used[vertex.index]) {
            // continue;
            // }
            // used[vertex.index] = true;

            if (vertex.cost > cost[vertex.index]) {
                continue;
            }
            // if (cost[vertex.index] < INF) {
            // Utils.debug("extract negative loop");
            // break;
            // }
            // cost[vertex.index] = vertex.cost;

            for (Edge edge : edges[vertex.index]) {
                // if (used[edge.to]) {
                // continue;
                // }
                long nextCost = vertex.cost + (b ? edge.yen : edge.snuke);
                if (nextCost >= cost[edge.to]) {
                    continue;
                }
                // assert edge.from == vertex.index;
                pq.add(new Vertex(edge.to, nextCost));
                cost[edge.to] = nextCost;
            }

        }
    }

}

class Edge {
    int to;
    long yen;
    long snuke;

    public Edge(int to, long y, long s) {
        super();
        this.to = to;
        this.yen = y;
        this.snuke = s;
    }
}

class Vertex {
    int index;
    long cost;

    public Vertex(int index, long cost) {
        super();
        this.index = index;
        this.cost = cost;
    }

}
