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;
using Pll = pair<ll, ll>;
vector<Pll> g[100005], h[100005];
ll n, m, s, t, u, v, a, b, dg[100005], dh[100005], ans[100005];

constexpr ll INF = (1LL << 60);

priority_queue<Pll, vector<Pll>, greater<Pll>> Q;

int main() {
  cin >> n >> m >> s >> t;
  s--, t--;
  for (int i = 0; i < m; ++i) {
    cin >> u >> v >> a >> b;
    u--, v--;
    g[u].push_back({v, a});
    g[v].push_back({u, a});
    h[u].push_back({v, b});
    h[v].push_back({u, b});
  }
  for (int i = 0; i < n; ++i) dg[i] = INF, dh[i] = INF;
  dg[s] = 0; Q.push({0, s});

  while (!Q.empty()) {
    int p = Q.top().second; Q.pop();
    for (const auto &f : g[p]) {
      ll cost = f.second, to = f.first;
      if (dg[to] > dg[p] + cost) {
        dg[to] = dg[p] + cost;
        Q.push({dg[to], to});
      }
    }
  }

  dh[t] = 0; Q.push({0, t});
  while (!Q.empty()) {
    int p = Q.top().second; Q.pop();
    for (const auto &f : h[p]) {
      ll cost = f.second, to = f.first;
      if (dh[to] > dh[p] + cost) {
        dh[to] = dh[p] + cost;
        Q.push({dh[to], to});
      }
    }
  }

  ans[n-1] = dg[n-1] + dh[n-1];
  for (int i = n-1; i > 0; --i) {
    ans[i-1] = min(ans[i], dg[i-1] + dh[i-1]);
  }
  for (int i = 0; i < n; ++i) {
    cout << 1000000000000000LL - ans[i] << endl;
  }
  return 0;
}
