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;

template<typename T> class Dijkstra {
public:
  struct edge {
    int to; T cost;
  };
  int V;
  vector<vector<edge>> G;
  vector<T> dist;
  using pti = pair<T,int>;
  Dijkstra(int n) : V(n), G(V), dist(V, numeric_limits<T>::max()) {}
  void add(int u, int v, T cost) {
    G.at(u).push_back((edge) {v, cost});
    G.at(v).push_back((edge) {u, cost});
  }
  void solve(int s) {
    priority_queue<pti, vector<pti>, greater<pti>> Q;
    dist.at(s) = 0;
    Q.push(pti(0, s));
    while (!Q.empty()) {
      pti p = Q.top();
      Q.pop();
      int v = p.second;
      if (dist.at(v) < p.first) continue;
      for (auto &w : G.at(v)) {
        if (dist.at(w.to) > dist.at(v) + w.cost) {
          dist.at(w.to) = dist.at(v) + w.cost;
          Q.push(pti(dist.at(w.to), w.to));
        }
      }
    }
  }
};

int main() {
  int N, M, S, G;
  cin >> N >> M >> S >> G;
  Dijkstra<long> D(N), D2(N);
  for (int i = 0; i < M; i++) {
    int a, b, c, d;
    cin >> a >> b >> c >> d;
    D.add(--a, --b, c);
    D2.add(a, b, d);
  }
  D.solve(--S);
  D2.solve(--G);
  vector<long> sum(N);
  for (int i = 0; i < N; i++) {
    sum.at(i) = D.dist.at(i) + D2.dist.at(i);
  }
  for (long i = N - 1, mi = LONG_MAX; i >= 0; i--) {
    sum.at(i) = (mi = min(mi, sum.at(i)));
  }
  for (int i = 0; i < N; i++) {
    cout << (long) 1e15 - sum.at(i) << "\n";
  }
}