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 rep(i,n) for (long long i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll,ll>;
using vec = vector<ll>;
using vecp = vector<P>;
using mat = vector<vec>;
using matp = vector<vecp>;
const ll MOD = 1e9+7;
const ll INF = 1e18;
#define all(v) v.begin(), v.end()

//matp G(N);
void graphmake(ll E,matp &GA,matp &GB){
  rep(i,E){
    ll a,b,c,d;
    cin >> a >> b >> c >> d;
    GA.at(a-1).push_back(P(b-1,c));
    GA.at(b-1).push_back(P(a-1,c));
    GB.at(a-1).push_back(P(b-1,d));
    GB.at(b-1).push_back(P(a-1,d));
  }
}
//vec d(N);
void dijkstra(ll s,vec &d, matp &G,ll V){
  priority_queue<P,vector<P>,greater<P>> que;
  rep(i,V) d.at(i)=INF;
  d.at(s)=0;
  que.push(P(0,s));
  while(!que.empty()){
    P p=que.top();
    que.pop();
    ll v=p.second;
    if(d.at(v)<p.first) continue;
    rep(i,G.at(v).size()){
      P e=G.at(v).at(i);
      if(d.at(e.first)>d.at(v)+e.second){
        d.at(e.first)=d.at(v)+e.second;
        que.push(P(d.at(e.first),e.first));
      }
    }
  }
}

int main(){
  ll N,M,s,t;
  cin >> N >> M >> s >> t;
  s--;
  t--;
  matp GA(N),GB(N);
  vec dA(N),dB(N);
  graphmake(M,GA,GB);
  dijkstra(s,dA,GA,N);
  dijkstra(t,dB,GB,N);
  priority_queue<P> Q;
  rep(i,N){
    Q.push(P(-dA.at(i)-dB.at(i),i));
  }
  rep(i,N){
    while(Q.top().second<i){
      Q.pop();
    }
    cout << 1000000000000000+Q.top().first << endl;
  }
}