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;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
const ll INF = 1LL << 60; //MAX 9223372036854775807

struct edge
{
  ll to;
  ll cost;
  edge() {}
  edge(ll to, ll cost) : to(to), cost(cost){};
};

vector<ll> dijkstra(ll s, vector<vector<edge>> &g)
{
  //pair.first=距離,pair.second=点
  priority_queue<pair<ll, ll>, vector<pair<ll, ll>>, greater<pair<ll, ll>>> q;
  vector<ll> dist(g.size(), INF);
  dist[s] = 0;
  q.push({0, s});

  while (!q.empty())
  {
    pair<ll, ll> p = q.top();
    q.pop();
    ll v = p.second;
    if (dist[v] < p.first)
      continue;
    rep(i, g[v].size())
    {
      edge e = g[v][i];
      if (dist[e.to] > dist[v] + e.cost)
      {
        dist[e.to] = dist[v] + e.cost;
        q.push({dist[e.to], e.to});
      }
    }
  }
  return dist;
}
int main()
{
ll n, m,s,t;
cin >> n >> m >> s >> t ;
vector<vector<edge>> g(n);
vector<vector<edge>> gg(n);
rep(i, m)
{
  ll a, b, c , d;
  cin >> a >> b >> c >> d;
  g[a - 1].emplace_back(b - 1, c);
  g[b - 1].emplace_back(a - 1, c);
  gg[a - 1].emplace_back(b - 1, d);
  gg[b - 1].emplace_back(a - 1, d);
}
auto v=dijkstra(s-1,g);
auto vv=dijkstra(t-1,gg);
vector<pair<ll,ll>> p;
rep(i,n){
  p.emplace_back(v[i]+vv[i],i);
}
sort(p.begin(),p.end());
ll itr=0;
rep(i,n){
  while(p[itr].second<i)itr++;
  cout << 1000000000000000-p[itr].first << endl;
  
}
return 0;
}