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;
using Int = long long;
 
template <typename T>
vector<T> dijkstra(Int s,vector<vector<pair<Int, T> > > & G,T INF){
  using P = pair<T, Int>;
  Int n=G.size();
  vector<T> d(n,INF);
  vector<Int> b(n,-1);
  priority_queue<P,vector<P>,greater<P> > q;
  d[s]=0;
  q.emplace(d[s],s);
  while(!q.empty()){
    P p=q.top();q.pop();
    Int v=p.second;
    if(d[v]<p.first) continue;
    for(auto& e:G[v]){
      Int u=e.first;
      T c=e.second;
      if(d[u]>d[v]+c){
	d[u]=d[v]+c;
	b[u]=v;
	q.emplace(d[u],u);
      }
    }
  }
  return d;
}
//INSERT ABOVE HERE
signed main(){
  Int n,m,s,t;
  cin>>n>>m>>s>>t;
  s--;t--;
  
  using P = pair<Int, Int>;
  vector<vector<P> > G(n),H(n);
  for(Int i=0;i<m;i++){
    Int x,y,a,b;
    cin>>x>>y>>a>>b;
    x--;y--;
    G[x].emplace_back(y,a);
    G[y].emplace_back(x,a);
    H[x].emplace_back(y,b);
    H[y].emplace_back(x,b);
  }
  
  const Int INF = 1e17;
  auto A=dijkstra(s,G,INF);
  auto B=dijkstra(t,H,INF);
 
  vector<Int> ans(n+1,INF);
  for(Int i=n-1;i>=0;i--)
    ans[i]=min(ans[i+1],A[i]+B[i]);
 
  Int tmp=1e15;
  for(Int i=0;i<n;i++) cout<<tmp-ans[i]<<endl;
  return 0;
}