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 rep(i, n) for (int i = 0; i < (int)(n); i++)

using P = pair<long long, int>;
const long long INF = 1e18 + 10;

struct Edge {
  int to, w;
  Edge(int t, int w) : to(t), w(w) {}
};

struct Graph {
  int N, M;
  vector<vector<Edge>> edges;

  Graph(int N, int M) : N(N), M(M), edges(N) {}

  void addEdge(int a, int b) { addEdge(a, b, 1); }
  void addEdge(int a, int b, int w) { edges[a].emplace_back(b, w); }

  vector<long long> dist;
  void dijkstra(int s) {
    dist = vector<long long>(N, INF);
    priority_queue<P, vector<P>, greater<P>> pq;
    pq.emplace(0, s);
    while (!pq.empty()) {
      long long d, now;
      tie(d, now) = pq.top();
      pq.pop();
      if (dist[now] != INF) continue;
      dist[now] = d;
      for (Edge e : edges[now]) {
        if (dist[e.to] == INF) pq.emplace(d + e.w, e.to);
      }
    }
  }
};

int main() {
  cout << setprecision(15);
  int n, m, s, t; cin >> n >> m >> s >> t;
  s--; t--;
  Graph Ga(n, m), Gb(n, m);
  rep(i, m) {
    int u, v, a, b; cin >> u >> v >> a >> b;
    u--; v--;
    Ga.addEdge(u, v, a);
    Ga.addEdge(v, u, a);
    Gb.addEdge(u, v, b);
    Gb.addEdge(v, u, b);
  }
  Ga.dijkstra(s);
  Gb.dijkstra(t);

  vector<P> ans(n);
  rep(i, n) ans[i] = P(Ga.dist[i] + Gb.dist[i], i);
  sort(ans.begin(), ans.end());
  int now = 0;
  rep(i, n) {
    while (ans[now].second < i) now++;
    cout << 1e15 - ans[now].first << endl;
  }
}