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 <iostream>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const long long INF = 1e18;

vector<long long> dijkstra(vector< vector< pair<long long, int> > > &G, int s){
  int n = G.size();
  vector<long long> ret(n, INF);
  priority_queue< pair<long long, int> > wait;
  wait.emplace(0,s);
  ret[s] = 0;
  while(!wait.empty()){
    int v = wait.top().second;
    long long cost = -wait.top().first;
    wait.pop();
    if(cost > ret[v]) continue;
    for(int i = 0; i < (int)G[v].size(); ++i){
      int v_ = G[v][i].second;
      long long cost_ = cost + G[v][i].first;
      if(cost_ < ret[v_]){
        ret[v_] = cost_;
        wait.emplace(-cost_,v_);
      }
    }
  }
  return ret;
}

int main(){
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  vector< vector< pair<long long, int> > > Ga(n), Gb(n);
  for(int i = 0; i < m; ++i){
    int u, v;
    long long a, b;
    cin >> u >> v >> a >> b;
    --u,--v;
    Ga[u].emplace_back(a,v);
    Ga[v].emplace_back(a,u);
    Gb[u].emplace_back(b,v);
    Gb[v].emplace_back(b,u);
  }
  vector<long long> A = dijkstra(Ga,s-1), B = dijkstra(Gb, t-1);
  vector<long long> ans(n,(long long)(1e15));
  long long d = INF;
  for(int i = n-1; i >= 0; --i){
    d = min(d, A[i]+B[i]);
    ans[i] -= d;
  }
  for(int i = 0; i < n; ++i) cout << ans[i] << endl;
  return 0;
}
