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 REPR(i, n) for (int i = n; i >= 0; i--)
#define FOR(i, m, n) for (int i = m; i < n; i++)
#define FORR(i, m, n) for (int i = m; i >= n; i--)
#define ALL(x) (x).begin(), (x).end()

using namespace std;
typedef long long ll;
typedef pair<ll, ll> P;
const ll INF = 1e15;

using edge = pair<ll, ll>;  // {to, cost}
using Graph = vector<vector<edge>>;

vector<ll> dijkstra(const Graph &g, ll s) {
  vector<ll> dist((int)g.size(), INF);
  priority_queue<P, vector<P>, greater<P>> que;
  dist[s] = 0;
  que.push(P(0, s));
  while (!que.empty()) {
    ll cost, idx;
    tie(cost, idx) = que.top();
    que.pop();
    if (dist[idx] < cost) continue;
    for (auto e : g[idx]) {
      if (dist[e.first] > cost + e.second) {
        dist[e.first] = cost + e.second;
        que.push(P(dist[e.first], e.first));
      }
    }
  }

  return dist;
}

int main() {
  ll n, m, s, t;
  cin >> n >> m >> s >> t;
  --s, --t;
  Graph g1(n), g2(n);
  REP(i, m) {
    ll u, v, a, b;
    cin >> u >> v >> a >> b;
    --u, --v;
    g1[u].push_back(edge(v, a));
    g1[v].push_back(edge(u, a));
    g2[u].push_back(edge(v, b));
    g2[v].push_back(edge(u, b));
  }

  auto d1 = dijkstra(g1, s);
  auto d2 = dijkstra(g2, t);

  vector<ll> ans;
  ll mn = INF;
  REPR(i, n - 1) {
    mn = min(mn, d1[i] + d2[i]);
    ans.push_back(mn);
  }

  reverse(ALL(ans));
  REP(i, n) { cout << INF - ans[i] << endl; }

  return 0;
}