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
#include<cstdio>
#include<utility>
#include<vector>
#include<queue>
#include<algorithm>
#define MAX_V 100001
#define INF 1000000000000000
using namespace std;
struct edge{int to;long cost;};
typedef pair<long,int> P;

int V;
vector<edge> G[MAX_V][2];
long d[MAX_V][2];

void dijkstra(int s,int b){
    priority_queue<P,vector<P>,greater<P> > que;
    for(int i=0;i<=V;i++){
        d[i][b]=INF;
    }
    d[s][b]=0;
    que.push(P(0,s));

    while(!que.empty()){
        P p=que.top(); que.pop();
        int v=p.second;
        if(d[v][b]<p.first) continue;
        for(int i=0;i<G[v][b].size();i++){
            edge e=G[v][b][i];
            if(d[e.to][b]>d[v][b]+e.cost){
                d[e.to][b]=d[v][b]+e.cost;
                que.push(P(d[e.to][b],e.to));
            }
        }
    }
}

int main(){
    int m,s,t,v,u;
    long a,b,ans[100000]={};
    scanf("%d%d%d%d",&V,&m,&s,&t);
    for(int i=0;i<m;i++){
        scanf("%d%d%ld%ld",&v,&u,&a,&b);
        G[u][0].push_back({v,a});
        G[u][1].push_back({v,b});
        G[v][0].push_back({u,a});
        G[v][1].push_back({u,b});
    }
    dijkstra(s,0);
    dijkstra(t,1);
    ans[V]=INF-(d[V][0]+d[V][1]);
    for(int i=V-1;i>=0;i--){
        ans[i]=INF-(d[i][0]+d[i][1]);
        ans[i]=max(ans[i],ans[i+1]);
    }
    for(int i=1;i<=V;i++){
        printf("%ld\n",ans[i]);
    }
}
