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;
typedef long long ll;

struct state {
  ll v, cost;
  bool operator>(const state s) const { return cost > s.cost; }
};

vector<ll> dijkstra(const vector<vector<pair<int, ll>>> &graph, int s) {
  vector<ll> dp(graph.size(), 1e18);
  priority_queue<state, vector<state>, greater<state>> q;
  q.push(state{s, 0});
  while (!q.empty()) {
    int v = q.top().v;
    ll cost = q.top().cost;
    q.pop();
    if (dp[v] <= cost) continue;
    dp[v] = cost;
    for (auto e : graph[v]) {
      ll nv = e.first, ncost = cost + e.second;
      if (dp[nv] > ncost) q.push(state{nv, ncost});
    }
  }
  return dp;
}

int main() {
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--;
  t--;
  vector<vector<pair<int, ll>>> a(n), b(n);
  for (int i = 0; i < m; i++) {
    int u, v, ai, bi;
    cin >> u >> v >> ai >> bi;
    u--;
    v--;
    a[u].push_back({v, ai});
    a[v].push_back({u, ai});
    b[u].push_back({v, bi});
    b[v].push_back({u, bi});
  }
  vector<ll> c = dijkstra(a, s), d = dijkstra(b, t);
  vector<ll> r(n);
  for (int i = 0; i < n; i++) {
    r[i] = c[i] + d[i];
  }
  for (int i = n - 2; i >= 0; i--) {
    r[i] = min(r[i], r[i + 1]);
  }
  for (auto e : r) {
    cout << (ll)1e15 - e << endl;
  }
}
