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<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define INF 1000000000000000000LL
typedef pair<int,int>Q;
typedef pair<long long,int>T;
typedef pair<Q,Q>P;
long long min_cost[5][100009];
vector<P>node[100009];

int main(){
  int n,m,s,t;
  cin>>n>>m>>s>>t;
  for(int j=0;j<2;j++){
    for(int i=0;i<=n;i++){
      min_cost[j][i]=INF;
    }
  }
  for(int i=0;i<m;i++){
    int u,v,a,b;
    cin>>u>>v>>a>>b;
    node[u].push_back(P(Q(v,a),Q(b,0)));
    node[v].push_back(P(Q(u,a),Q(b,0)));
  }
  for(int j=0;j<2;j++){
    priority_queue<T,vector<T>,greater<T> >pq;
    if(j==0){
      pq.push(T(0,s));
      min_cost[0][s]=0;
    }
    else{
      pq.push(T(0,t));
      min_cost[1][t]=0;
    }
    while(!pq.empty()){
      T now=pq.top();pq.pop();
      long long cost=now.fi;
      int pos=now.se;
      for(int i=0;i<node[pos].size();i++){
        int next=node[pos][i].fi.fi;
        long long ncost;
        if(j==0)ncost=(long long)node[pos][i].fi.se;
        else ncost=(long long)node[pos][i].se.fi;
        if(cost+ncost<min_cost[j][next]){
          min_cost[j][next]=cost+ncost;
          pq.push(T(cost+ncost,next));
        }
      }
    }
  }
  priority_queue<T,vector<T>,greater<T> >p;
  for(int i=1;i<=n;i++){
    p.push(T(min_cost[0][i]+min_cost[1][i],i));
  }
  for(int i=0;i<n;i++){
    T now;
    while(1){
      now=p.top();
      if(now.se<=i)p.pop();
      else break;
    }
    cout<<1000000000000000LL-now.fi<<endl;
  }
  return(0);
}