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>
#define rep(i, n) for (int i = 0; i < (n); ++i)
#define repI(i, d, n) for (int i = (d); i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<ll, ll>;
using PP = pair<ll, P>;
using M = map<ll, ll>;
ll INF = 100100100100100;
ll MONEY = 1e15;

vector<ll> dijkstra(ll start, vector<P> *G, ll n) {
  priority_queue<P, vector<P>, greater<P>> pq;
  pq.push(P(0, start));

  vector<ll> dist(n);
  rep(i, n) dist[i] = INF;
  dist[start] =  0;

  while (! pq.empty()) {
    ll pos = pq.top().second;
    pq.pop();

    rep(i, int(G[pos].size())) {
      ll cost = G[pos][i].second;
      ll to = G[pos][i].first;

      if (dist[to] > dist[pos] + cost) {
        dist[to] = dist[pos] + cost;
        pq.push(P(dist[to], to));
      }
    }
  }

  return dist;
}

int main() {
  ll n, m, s, t;
  cin >> n >> m >> s >> t;
  s--; t--;

  vector<P> G_s[n];
  vector<P> G_t[n];

  rep(i, m) {
    ll u, v, a, b;
    cin >> u >> v >> a >> b;
    u--; v--;
    G_s[u].push_back(P(v, a));
    G_s[v].push_back(P(u, a));
    G_t[u].push_back(P(v, b));
    G_t[v].push_back(P(u, b));
  }

  vector <ll> dist_s = dijkstra(s, G_s, n);
  vector <ll> dist_t = dijkstra(t, G_t, n);

  vector<ll> dist(n);
  rep(i, n) {
    dist[i] = dist_s[i] + dist_t[i];
  }

  vector<ll> ans(n);
  ans[n-1] = dist[n-1];

  rep(i, n-1) {
    ans[n-2-i] = min(dist[n-2-i], ans[n-1-i]);
  }

  rep(i, n) {
    cout << MONEY - ans[i] << endl;
  }

  return 0;
}
