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.List;
import java.util.PriorityQueue;
import java.util.Scanner;

class Main {
    public static class Dijkstra {
        Vertex[] g;
        PriorityQueue<Pair> que;
        long[] d;

        public Dijkstra(int n) {
            this.g = new Vertex[n];
            this.d = new long[n];
            this.que = new PriorityQueue<>();

            for (int i = 0; i < n; i++) {
                this.g[i] = new Vertex(i);
            }
            Arrays.fill(this.d, Long.MAX_VALUE / 4);
        }

        public void set(int i, int j, long cost) {
            Vertex u = this.g[i];
            Vertex v = this.g[j];
            u.adj.add(new Edge(cost, v));
            v.adj.add(new Edge(cost, u));
        }

        public void calc(int s) {
            this.d[s] = 0;
            this.que.add(new Pair(0, this.g[s]));

            while (this.que.size() > 0) {
                Pair t = this.que.poll();
                int from = t.now.name;
                List<Edge> toes = t.now.adj;
                if (this.d[from] < t.value) {
                    continue;
                }
                for (Edge e : toes) {
                    int to = e.to.name;
                    long cost = e.w;
                    if (this.d[to] > this.d[from] + cost) {
                        this.d[to] = this.d[from] + cost;
                        this.que.add(new Pair(this.d[to], e.to));
                    }
                }
            }
        }

        public long distance(int g) {
            return this.d[g];
        }
    }

    public static class Edge {
        long w;
        Vertex to;

        public Edge(long w, Vertex to) {
            this.w = w;
            this.to = to;
        }

    }

    public static class Vertex {
        int name;
        List<Edge> adj;

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

    }

    public static class Pair implements Comparable<Pair> {
        long value;
        Vertex now;

        public Pair(long value, Vertex now) {
            this.value = value;
            this.now = now;
        }

        @Override
        public int compareTo(Pair o) {

            return Long.compare(this.value, o.value);
        }

    }

    public static void main(String[] args) {
        Scanner scanner = new Scanner(System.in);
        int N = scanner.nextInt();
        int M = scanner.nextInt();
        int s = scanner.nextInt() - 1;
        int t = scanner.nextInt() - 1;
        Dijkstra sDijkstra = new Dijkstra(N);
        Dijkstra tDijkstra = new Dijkstra(N);
        for (int i = 0; i < M; i++) {
            int v = scanner.nextInt() - 1;
            int u = scanner.nextInt() - 1;
            long a = scanner.nextLong();
            long b = scanner.nextLong();
            sDijkstra.set(v, u, a);
            tDijkstra.set(v, u, b);
        }
        sDijkstra.calc(s);
        tDijkstra.calc(t);
        long[] ans = new long[N];
        ans[N - 1] = (long) Math.pow(10, 15) - sDijkstra.distance(N - 1) - tDijkstra.distance(N - 1);
        for (int i = N - 2; i >= 0; i--) {
            ans[i] = Math.max(ans[i + 1], (long) Math.pow(10, 15) - sDijkstra.distance(i) - tDijkstra.distance(i));
        }
        for (long l : ans) {
            System.out.println(l);
        }
    }
}
