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>
#include <assert.h>

using namespace std;

struct Edge
{
  long long to, cost;
};

void dijkstra(vector<Edge> edge[], int n, int s, vector<long long> &far)
{
  priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
  far[s] = 0;
  pq.push({0, s});
  while (pq.size())
  {
    pair<long long, int> p = pq.top();
    pq.pop();
    cerr << " " << p.second << " " << p.first << endl;
    for (Edge t : edge[p.second])
      if (far[p.second] + t.cost < far[t.to])
      {
        far[t.to] = far[p.second] + t.cost;
        pq.push({far[t.to], t.to});
      }
  }
}

int main()
{
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  vector<Edge> edgeY[n], edgeS[n];
  for (int i = 0; i < m; i++)
  {
    long long u, v, a, b;
    cin >> u >> v >> a >> b;
    u--;
    v--;
    edgeY[u].push_back({v, a});
    edgeY[v].push_back({u, a});
    edgeS[u].push_back({v, b});
    edgeS[v].push_back({u, b});
  }
  vector<long long> costY(n, 1e18), costS(n, 1e18);
  dijkstra(edgeY, n, s - 1, costY);
  dijkstra(edgeS, n, t - 1, costS);
  vector<pair<long long, int>> cost(n);
  for (int i = 0; i < n; i++)
    cost[i] = {costY[i] + costS[i], i + 1};
  sort(cost.begin(), cost.end());
  int index = 0;
  for (int i = 0; i < n; i++)
  {
    while (cost[index].second <= i)
      index++;
    cout << (long long)1e15 - cost[index].first << endl;
  }

  return 0;
}
