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 lint = long long;
using VI = vector<int>;
using VVI = vector<VI>;
using P = pair<lint, int>;

const lint INF = 1e18;

int main() {
  cin.tie(0); ios::sync_with_stdio(false);
  int n, m, s, t; cin >> n >> m >> s >> t;
  s--; t--;
  struct edge { int to, y, s; };
  vector< vector<edge> > G(n, vector<edge>());
  for (int i = 0; i < m; i++) {
    int u, v, a, b; cin >> u >> v >> a >> b;
    u--; v--;
    G[u].push_back(edge{v, a, b});
    G[v].push_back(edge{u, a, b});
  }
  
  vector<lint> dy(n, INF), ds(n, INF);
  queue<P> pq;
  dy[s] = 0;
  pq.push(P(0, s));
  while (not pq.empty()) {
    P p = pq.front(); pq.pop();
    int v = p.second;
    if (dy[v] < p.first) continue;
    for (auto&& e : G[v]) {
      if (dy[v] + e.y < dy[e.to]) {
        dy[e.to] = dy[v] + e.y;
        pq.push(P(dy[e.to], e.to));
      }
    }
  }

  ds[t] = 0;
  pq.push(P(0, t));
  while (not pq.empty()) {
    P p = pq.front(); pq.pop();
    int v = p.second;
    if (ds[v] < p.first) continue;
    for (auto&& e : G[v]) {
      if (ds[v] + e.s < ds[e.to]) {
        ds[e.to] = ds[v] + e.s;
        pq.push(P(ds[e.to], e.to));
      }
    }
  }

  vector<lint> res(n);
  const lint N = 1e15;
  for (int i = 0; i < n; i++) {
    res[i] = N - dy[i] - ds[i];
  }
  for (int i = n - 2; i >= 0; i--) {
    res[i] = max(res[i], res[i + 1]);
  }
  for (int i = 0; i < n; i++) {
    cout << res[i] << '\n';
  }
  return 0;
}
