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 ll=long long;
#define MAX 300003
#define MOD 998244353
#define INF 1000000000000000

int main(){
  int n,m,s,t;
  cin>>n>>m>>s>>t;
  s--,t--;
  vector<vector<pair<int,int>>> G(n);
  vector<ll> a(m),b(m);
  for(int i=0;i<m;i++){
    int u,v;
    cin>>u>>v>>a.at(i)>>b.at(i);
    u--;v--;
    G.at(u).push_back(make_pair(v,i));
    G.at(v).push_back(make_pair(u,i));
  }

  vector<ll> cost_y(n,INF);
  cost_y.at(s)=0;
  priority_queue<pair<ll,int>,vector<pair<ll,int>>,greater<pair<ll,int>>> pq;
  pq.push(make_pair(cost_y.at(s),s));
  while(!pq.empty()){
    int v=pq.top().second;
    pq.pop();
    for(auto p:G.at(v)){
      int nv=p.first;
      int i=p.second;
      if(cost_y.at(nv)>cost_y.at(v)+a.at(i)){
        cost_y.at(nv)=cost_y.at(v)+a.at(i);
        pq.push(make_pair(cost_y.at(nv),nv));
      }
    }
  }

  vector<ll> cost_s(n,INF);
  cost_s.at(t)=0;
  pq.push(make_pair(cost_s.at(t),t));
  while(!pq.empty()){
    int v=pq.top().second;
    pq.pop();
    for(auto p:G.at(v)){
      int nv=p.first;
      int i=p.second;
      if(cost_s.at(nv)>cost_s.at(v)+b.at(i)){
        cost_s.at(nv)=cost_s.at(v)+b.at(i);
        pq.push(make_pair(cost_s.at(nv),nv));
      }
    }
  }

  vector<ll> ans(n);
  ans.at(n-1)=cost_y.at(n-1)+cost_s.at(n-1);
  for(int i=n-2;i>=0;i--){
    ans.at(i)=min<ll>(ans.at(i+1),cost_y.at(i)+cost_s.at(i));
  }

  for(int i=0;i<n;i++){
    cout<<1000000000000000-ans.at(i)<<endl;
  }
}
