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;

int64_t I=167167167167167;

int main(){
  int N,M,S,T;
  cin>>N>>M>>S>>T;
  S--,T--;
  vector<vector<tuple<int,int64_t,int64_t>>> G(N);
  for(int i=0;i<M;i++){
    int a,b;
    int64_t c,d;
    cin>>a>>b>>c>>d;
    a--,b--;
    G[a].push_back(make_tuple(b,c,d));
    G[b].push_back(make_tuple(a,c,d));
  }
  vector<int64_t> yen(N,I);
  vector<int64_t> snuke(N,I);
  priority_queue<pair<int64_t,int>> pq;
  pq.push(make_pair(0,S));
  yen[S]=0;
  while(!pq.empty()){
    int64_t a;
    int b;
    tie(a,b)=pq.top();
    pq.pop();
    a*=-1;
    if(yen[b]<a){
      continue;
    }
    for(auto x:G[b]){
      int c;
      int64_t d,e;
      tie(c,d,e)=x;
      if(yen[c]>d+a){
        yen[c]=d+a;
        pq.push(make_pair((d+a)*-1,c));
      }
    }
  }
  pq.push(make_pair(0,T));
  snuke[T]=0;
  while(!pq.empty()){
    int64_t a;
    int b;
    tie(a,b)=pq.top();
    pq.pop();
    a*=-1;
    if(snuke[b]<a){
      continue;
    }
    for(auto x:G[b]){
      int c;
      int64_t d,e;
      tie(c,d,e)=x;
      if(snuke[c]>e+a){
        snuke[c]=e+a;
        pq.push(make_pair((e+a)*-1,c));
      }
    }
  }
  vector<int64_t> p(N,1167167167167167);
  for(int i=0;i<N;i++){
    p[i]-=I;
    p[i]-=yen[i];
    p[i]-=snuke[i];
  }
  for(int i=N-1;i>0;i--){
    p[i-1]=max(p[i],p[i-1]);
  }
  for(int i=0;i<N;i++){
    cout<<p[i]<<endl;
  }
  
}