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 ll = long long;

const int N = 1e5;
vector<pair<int, ll>> g[2][N];
int n, m, s, t;
ll inf = 1e18;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cin >> n >> m >> s >> t;
  s--; t--;
  for(int i = 0; i < m; i++) {
    ll a, b, x, y; cin >> a >> b >> x >> y;
    a--; b--;
    g[0][a].emplace_back(b, x);
    g[0][b].emplace_back(a, x);
    g[1][a].emplace_back(b, y);
    g[1][b].emplace_back(a, y);
  }
  // dijkstra
  using P = pair<ll, int>;
  priority_queue<P, vector<P>, greater<P>> pq;
  vector<vector<ll>> dists(2, vector<ll>(n, inf));
  for(int p = 0; p < 2; p++) {
    auto & dist = dists[p];
    pq.emplace(0, p ? t : s);
    for(int i = 0; i < n; i++) dist[i] = inf;
    dist[p ? t : s] = 0;
    while(pq.size()) {
      int i; ll di;
      tie(di, i) = pq.top();
      pq.pop();
      if(dist[i] < di) continue;
      for(auto to : g[p][i]) {
        ll j, cost;
        tie(j, cost) = to;
        if(dist[j] > di + cost) {
          dist[j] = di + cost;
          pq.emplace(di + cost, j);
        }
      }
    }
  }
  vector<ll> ans(n+1);
  ans[n] = inf;
  for(int i = n-1; i >=0; i--) {
    ans[i] = min(ans[i+1], dists[0][i] + dists[1][i]);
  }
  for(int i = 0; i < n; i++) cout << ll(1e15) - ans[i] << endl;
  return 0;
}
