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{
    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;
        
        Dijkstra en = new Dijkstra(n);
        Dijkstra sn = new Dijkstra(n);
        
        for(int i=0; i<m; i++){
            int u = Integer.parseInt(sc.next())-1;
            int v = Integer.parseInt(sc.next())-1;
            int a = Integer.parseInt(sc.next());
            int b = Integer.parseInt(sc.next());
            
            en.addEdge(u, v, a);
            sn.addEdge(u, v, b);
        }
        
        long[] ends = en.dijkstra(s);
        long[] snds = sn.dijkstra(t);
        
        long[] cost = new long[n];
        for(int i=0; i<n; i++){
            cost[i] = ends[i] + snds[i];
        }
        for(int i=n-1; i>0; i--){
            cost[i-1] = Math.min(cost[i-1], cost[i]);
        }
        for(int i=0; i<n; i++){
            System.out.println((long)1e15 - cost[i]);
        }
    }
}


class Dijkstra{
    
    public final long INF = (long)1e18;
    private int n;
    private List<Edge>[] edge;
    
    public Dijkstra(int n){
        this.n = n;
        edge = new List[n];
        for(int i=0; i<n; i++){
            edge[i] = new ArrayList<>();
        }
    }
    
    public void addEdge(int from, int to, int cost){
        edge[from].add(new Edge(from, to, cost));
        edge[to].add(new Edge(to, from, cost));
    }
    
    public long[] dijkstra(int from){
        long[] dist = new long[n];
        Arrays.fill(dist, INF);
        dist[from] = 0;
        
        PriorityQueue<Vertex> q = new PriorityQueue<>();
        q.add(new Vertex(from, 0));
        while(q.size() != 0){
            int u = q.poll().ver;
            
            for(Edge ne : edge[u]){
                int v = ne.to;
                long weight = ne.cost;
                if(dist[v] > dist[u]+weight){
                    dist[v] = dist[u]+weight;
                    q.add(new Vertex(v, dist[v]));
                }
            }
        }
        
        return dist;
    }
}

class Edge implements Comparable<Edge>{
    public int from = 0;
    public int to = 0;
    public int cost = 0;
    
    public Edge(int from, int to, int cost) {
        this.from = from;
        this.to = to;
        this.cost = cost;
    }
    
    @Override
    public int compareTo(Edge o) {
        return this.cost - o.cost;
    }
}

class Vertex implements Comparable<Vertex>{
    public int ver = 0;
    public long cost = 0;
    
    public Vertex(int ver, long cost) {
        this.ver = ver;
        this.cost = cost;
    }
    
    @Override
    public int compareTo(Vertex o) {
        long tmp = this.cost - o.cost;
        if(tmp == 0) return 0;
        return (int)(tmp/Math.abs(tmp));
    }
}