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>
#define int long long
using namespace std;
const int INF=1e16;
const int Ah=1e15;

signed main() {
  int N,M,S,T;cin>>N>>M>>S>>T;S--;T--;
  typedef pair<int,int> P;
  std::vector<P> en[N],su[N];
  for(int i=0;i<M;i++){
    int a,b,c,d;cin>>a>>b>>c>>d;a--;b--;
    en[a].push_back(P(b,c));
    en[b].push_back(P(a,c));
    su[a].push_back(P(b,d));
    su[b].push_back(P(a,d));
  }
  std::vector<int> de(N,INF),ds(N,INF);
  de[S]=0;ds[T]=0;
  priority_queue<P,vector<P>,greater<P>> que;
  que.push(P(0,S));
  while(que.size()){
    P p=que.top();que.pop();
    if(de[p.second]<p.first)continue;
    for(int i=0;i<en[p.second].size();i++){
      if(de[p.second]+en[p.second][i].second<de[en[p.second][i].first]){
        de[en[p.second][i].first]=de[p.second]+en[p.second][i].second;
        que.push(P(de[en[p.second][i].first],en[p.second][i].first));
      }
    }
  }
  que.push(P(0,T));
  while(que.size()){
    P p=que.top();que.pop();
    if(ds[p.second]<p.first)continue;
    for(int i=0;i<su[p.second].size();i++){
      if(ds[p.second]+su[p.second][i].second<ds[su[p.second][i].first]){
        ds[su[p.second][i].first]=ds[p.second]+su[p.second][i].second;
        que.push(P(ds[su[p.second][i].first],su[p.second][i].first));
      }
    }
  }
  std::vector<int> v(N);
  for(int i=0;i<N;i++)v[i]=de[i]+ds[i];
  std::vector<int> ans(N);
  ans[N-1]=v[N-1];
  for(int i=N-2;i>=0;i--)ans[i]=min(ans[i+1],v[i]);
  for(int i=0;i<N;i++)cout<<Ah-ans[i]<<endl;
}
