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
//Dijkstra-标程
#include <bits/stdc++.h>
#define ll long long
#define REP(i, n) for (ll (i) = 0; (i) < (n); (i)++)
#define REPI(i, a, b) for (ll (i) = (a); (i) < (b); (i)++)
#define VI vector<ll>
#define int long long
using namespace std;
 
#define INF 1e18
using P = pair<int, int>; // (distance, v)
struct edge {
  int to;
  int cost;
};
vector<int> dijkstra(vector<vector<edge>>& G, int start) {
  priority_queue<P, vector<P>, greater<P>> que;
  vector<int> d = vector<int>(G.size(), INF);
  d[start] = 0;
  que.push(P(0, start));
 
  while (!que.empty()) {
    P p = que.top(); que.pop();
    int v = p.second;
    if (d[v] < p.first) { continue; }
    for (int i = 0; i < G[v].size(); i++) {
      edge e = G[v][i];
      if (d[v] + e.cost < d[e.to]) {
        d[e.to] = d[v] + e.cost;
        que.push(P(d[e.to], e.to));
      }
    }
  }
  return d;
}
 
signed main() {
  int money = 1e15;
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--; t--;
  vector<vector<edge>> G1(n);
  vector<vector<edge>> G2(n);
  REP (i, m) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--; v--;
    G1[u].push_back({v, a});
    G1[v].push_back({u, a});
    G2[u].push_back({v, b});
    G2[v].push_back({u, b});
  }
 
  vector<int> d1(dijkstra(G1, s));
  vector<int> d2(dijkstra(G2, t));
 
  vector<int> ans(n+1);
  ans[n] = 0;
  for (int i = n-1; i >= 0; i--) {
    ans[i] = max(ans[i+1], money - (d1[i] + d2[i]));
  }
 
  REP (i, n) {
    cout << ans[i] << endl;
  }
}