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;
using P = pair<ll, int>;
const ll INF = 1e15;

struct Edge {
  int to;
  ll dist;
  Edge(int to, ll dist) : to(to), dist(dist) {}
};

struct Dijkstra {
  vector<ll> d;
  
  Dijkstra(int s, const vector<vector<Edge>> &g) {
    d.resize(g.size(), INF);
    d.at(s) = 0;
    
    priority_queue<P, vector<P>, greater<P>> que;
    que.push(P(0, s));
    
    while (!que.empty()) {
      P p = que.top();
      que.pop();
      int v = p.second;
      if (d.at(v) < p.first) continue;
      
      for (Edge e : g.at(v)) {
        if (d.at(e.to) > d.at(v) + e.dist) {
          d.at(e.to) = d.at(v) + e.dist;
          que.push(P(d.at(e.to), e.to));
        }
      }
    }
  }  
};

int main() {
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--;
  t--;
  
  vector<vector<Edge>> G(n); // 円の計算用
  vector<vector<Edge>> G2(n); // スヌークの計算用
  
  for (int i = 0; i < m; i++) {
    int u, v;
    ll a, b;
    cin >> u >> v >> a >> b;
    u--;
    v--;
    
    G.at(u).push_back(Edge(v, a));
    G.at(v).push_back(Edge(u, a));
    G2.at(u).push_back(Edge(v, b));
    G2.at(v).push_back(Edge(u, b));
  }
  
  Dijkstra di(s, G);
  Dijkstra di2(t, G2);
  
  vector<ll> ans(n + 1, INF);
  for (int i = n - 1; i >= 0; i--) {
    ans.at(i) = min(ans.at(i + 1), di.d.at(i) + di2.d.at(i));
  }
  
  ll tmp = 1e15;
  for (int i = 0; i < n; i++) {
    cout << tmp - ans.at(i) << endl;
  }
  
  return 0;
}