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.*;

public class Main {
    static int n;
    static int m;
    static int s;
    static int t;
    static long arigane = (long) 1e15;
    //    static int[][] grapht;
//    static int[][] graphf;
    static long[] tmemo;
    static long[] fmemo;
    static PriorityQueue<Path> paths;
    static long min = Long.MAX_VALUE;
    static List<Long> ret = new ArrayList<>();
    static Map<Integer, List<Edge>> graph = new HashMap<>();

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);

        n = sc.nextInt();
        m = sc.nextInt();
        s = sc.nextInt() - 1;
        t = sc.nextInt() - 1;
//        grapht = new int[n][n];
//        graphf = new int[n][n];
        tmemo = new long[n];
        fmemo = new long[n];

        for (int i = 0; i < n; i++) {
            tmemo[i] = -1;
            fmemo[i] = -1;
        }
        for (int i = 0; i < n; i++) {
            graph.put(i, new ArrayList<>());
        }
        for (int i = 0; i < m; i++) {
            int u = sc.nextInt() - 1;
            int v = sc.nextInt() - 1;
            int a = sc.nextInt();
            int b = sc.nextInt();
            Edge e1 = new Edge(v, a, b);
            Edge e2 = new Edge(u, a, b);
            graph.get(u).add(e1);
            graph.get(v).add(e2);

//            grapht[u][v] = a;
//            grapht[v][u] = a;
//            graphf[u][v] = b;
//            graphf[v][u] = b;
        }

        paths = new PriorityQueue<>(Comparator.comparingLong(x -> x.score));
        paths.add(new Path(s, 0));
        while (!paths.isEmpty()) {
            Path c = paths.poll();
            if (tmemo[c.current] != -1) continue;
            tmemo[c.current] = c.score;
//            for (int i = 0; i < n; i++) {
//                if(grapht[c.current][i] > 0 && tmemo[i] == -1) {
//                    Path tmp = new Path(i, c.score + grapht[c.current][i]);
//                    paths.add(tmp);
//                }
//            }
            for (Edge e : graph.get(c.current)) {
                if (tmemo[e.to] == -1) {
                    Path tmp = new Path(e.to, c.score + e.a);
                    paths.add(tmp);
                }
            }
        }

        paths.add(new Path(t, 0));
        while (!paths.isEmpty()) {
            Path c = paths.poll();
            if (fmemo[c.current] != -1) continue;
            fmemo[c.current] = c.score;
//                for (int i = 0; i < n; i++) {
//                    if (graphf[c.current][i] > 0 && fmemo[i] == -1) {
//                        Path tmp = new Path(i, c.score + graphf[c.current][i]);
//                        paths.add(tmp);
//                    }
//                }
            for (Edge e : graph.get(c.current)) {
                if (fmemo[e.to] == -1) {
                    Path tmp = new Path(e.to, c.score + e.b);
                    paths.add(tmp);
                }
            }
        }

        for (int i = 0; i < n; i++) {
            min = Math.min(min, tmemo[n - 1 - i] + fmemo[n - 1 - i]);
            ret.add(arigane - min);
        }

        for (int i = 0; i < n; i++) {
            System.out.println(ret.get(n - 1 - i));
        }

    }

    static class Path {
        int current;
        long score;

        Path(int current, long score) {
            this.current = current;
            this.score = score;
        }
    }

    static class Edge {
        int to;
        int a;
        int b;

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

}
