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

public class Main {

    static long iniPos = (long) Math.pow(10, 15);
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int n = sc.nextInt();
        int m = sc.nextInt();
        int s = sc.nextInt() -1;
        int t = sc.nextInt() -1;
        int[][] yenEdges = new int[m][3];//{from, to, cost}
        int[][] snuukEdges = new int[m][3];
        for (int i = 0; i < m; i++) {
            int from = sc.nextInt() -1;
            int to = sc.nextInt() -1;
            int a = sc.nextInt(); // yen
            int b = sc.nextInt(); // snuuk
            yenEdges[i] = new int[]{from, to, a};
            snuukEdges[i] = new int[]{from, to, b};
        }
        int[][][] yenGraph = buildWeightedGraph(n, yenEdges);
        int[][][] snuukGraph = buildWeightedGraph(n, snuukEdges);
        long[] yenDis = dijkstra(n, yenGraph, s);
        long[] snuukDis = dijkstra(n, snuukGraph, t);
        // 各頂点で両替した時の最小コストを降順に保持する {cost, vertex}
        PriorityQueue<long[]> q =
                new PriorityQueue<long[]>((x, y) -> Long.compare(x[0], y[0]));
        for (int i = 0; i < n; i++) {
            long cost = yenDis[i] + snuukDis[i];
            q.add(new long[]{cost, i});
        }
        StringBuilder sb = new StringBuilder();
        long[] a = q.peek();
        for (int i = 0; i < n; i++) {
            while(a[1] + 1 <= i){
                q.poll();
                a = q.peek();
            }
            sb.append(iniPos - a[0]).append("\n");
        }

        System.out.println(sb.toString());
        sc.close();


    }

    public static int[][][] buildWeightedGraph(int n, int[][] edges){
        int m = edges.length;
        int[][][] graph = new int[n][][];
        int[] deg = new int[n];
        for (int i = 0; i < m; i++) {
            deg[ edges[i][0] ]++;
            deg[ edges[i][1] ]++;
        }
        for (int i = 0; i < n; i++) {
            graph[i] = new int[deg[i]][2];
        }
        for (int i = 0; i < m; i++) {
            int a = edges[i][0];
            int b = edges[i][1];
            int c = edges[i][2];
            graph[a][--deg[a]][0] = b;
            graph[a][deg[a]][1] = c;
            graph[b][--deg[b]][0] = a;
            graph[b][deg[b]][1] = c;
        }

        return graph;
    }

    public static long[] dijkstra(int n, int[][][] graph, int s) {
        // {sからの最短距離, 頂点}
        PriorityQueue<long[]> pq =
                new PriorityQueue<long[]>((x, y) -> Long.compare(x[0], y[0]));
        long[] d = new long[n]; // sからの最短距離
        Arrays.fill(d, Long.MAX_VALUE);
        d[s] = 0;
        pq.add(new long[]{0, s});
        while(!pq.isEmpty()){
            long[] a = pq.poll();
            int v = (int)a[1];
            if(d[v] < a[0]) continue;
            for (int[] g: graph[v]) {
                int to = g[0];
                int cost = g[1];
                if(d[to] > d[v] + cost){
                    d[to] = d[v] + cost;
                    pq.add(new long[]{d[to], to});
                }
            }
        }

        return d;
    }

}
