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;
#define int long long
typedef pair<int, int> P;
const int INF = 1e16;
struct  edge{int to, cost;};
using Graph = vector<vector<edge> >;

void dij(Graph& G, int start, vector<int> &dist){
  priority_queue<P, vector<P>, greater<P> > que;
  for(int i = 0; i < dist.size(); i++){
    dist[i] = INF;
  }
  dist[start] = 0;
  que.push(P(0, start));
  
  while(!que.empty()){
    P p = que.top(); que.pop();
    int next = p.second;
    if(dist[next] < p.first) continue;
    for(int i = 0; i < G[next].size(); i++){
      edge e = G[next][i];
      if(dist[next] + e.cost < dist[e.to]){
        dist[e.to] = dist[next] + e.cost;
        que.push(P(dist[e.to], e.to));
      }
    }
  }
}

signed main(){
  int n, m, s, t; cin >> n >> m >> s >> t; --s; --t;
  Graph Gy(n), Gs(n);
  for(int i = 0; i < m; i++){
      int u, v, a, b; cin >> u >> v >> a >> b;
      --u; --v;
      edge e1 = {u, a}, e2 = {u, b},  e3 = {v, a}, e4 = {v, b};
      Gy[v].push_back(e1), Gs[v].push_back(e2), Gy[u].push_back(e3);
      Gs[u].push_back(e4); 
  }
  vector<int> money(n), dists(n), disty(n);
  dij(Gy, s, disty);
  dij(Gs, t, dists);
  for(int i = 0; i < n; i++) money[i] = dists[i] + disty[i];
  vector<int> min_money(n);
  min_money[n - 1] = money[n - 1];
  for(int i = n - 2; i >= 0; i--){
      min_money[i] = min(min_money[i + 1], money[i]);
  }
  for(int i = 0; i < n; i++){
      cout << 1000000000000000 - min_money[i] << endl;
  }
}