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 <algorithm>
#include <iostream>
#include <queue>
#include <vector>

using namespace std;

typedef long long ll;

#define MOD 1000000007

struct edge {
  int to;
  ll cost;
};
typedef pair<ll, ll> pl;
typedef vector<pl> vp;
typedef greater<pl> gp;

int n, m, s, t;
vector<edge> g[100000];
vector<edge> h[100000];

vector<ll> dijkstra(vector<edge> g[], int s) {
  priority_queue<pl, vp, gp> que;
  vector<ll> d(n, 1e18);
  d[s] = 0;
  que.push(make_pair(0, s));
  while (que.size()) {
    pl p = que.top();
    que.pop();
    int v = p.second;
    if (d[v] < p.first) continue;
    for (int i = 0; i < g[v].size(); ++i) {
      edge e = g[v][i];
      if (d[v] + e.cost < d[e.to]) {
        d[e.to] = d[v] + e.cost;
        que.push(make_pair(d[e.to], e.to));
      }
    }
  }
  return d;
}

int main() {
  cin >> n >> m >> s >> t;
  s--, t--;
  while (m--) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--, v--;
    edge e1 = {v, a}, e2 = {u, a};
    g[u].push_back(e1), g[v].push_back(e2);
    edge e3 = {v, b}, e4 = {u, b};
    h[u].push_back(e3), h[v].push_back(e4);
  }
  vector<ll> dist1 = dijkstra(g, s);
  vector<ll> dist2 = dijkstra(h, t);
  vector<ll> ans(n);
  for (int i = 0; i < n; ++i) {
    ans[i] = dist1[i] + dist2[i];
  }
  for (int i = n - 2; i >= 0; --i) {
    ans[i] = min(ans[i], ans[i + 1]);
  }
  ll x = 1e15;
  for (int i = 0; i < n; ++i) {
    cout << x - ans[i] << endl;
  }
  return 0;
}