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 ld = long double;
using pint = pair<int,  int>;
using pll = pair<ll, ll>;
using edge = pair<int, ll>;
using Graph = vector<vector<int>>; 
using WeightedGraph = vector<vector<edge>>;
const ll INF = 1LL << 60;
const int INF32 = 1 << 29;
const ll MOD = 1000000007;

vector<ll> dijkstra(int s,WeightedGraph  & G){
  using P = pair<ll, int>;
  int n=G.size();
  vector<ll> ds(n,INF);
  vector<int> bs(n,-1);
  priority_queue<P, vector<P>, greater<P> > pq;
  ds[s]=0;
  pq.emplace(ds[s],s);
  while(!pq.empty()){
    P p=pq.top();pq.pop();
    int v=p.second;
    if(ds[v]<p.first) continue;
    for(auto& e:G[v]){
      int u=e.first;
      ll c=e.second;
      if(ds[u]>ds[v]+c){
        ds[u]=ds[v]+c;
        bs[u]=v;
        pq.emplace(ds[u],u);
      }
    }
  }
  return ds;
}


int main(){
  int n,m,s,t;
  cin >> n >> m >> s >> t;
  --s;--t;
  
  WeightedGraph G(n), G2(n);
  for(int i = 0; i < m; ++i){
    ll u,v,a,b;
    cin >> u >> v >> a >> b;
    --u;--v;
    G[u].push_back({v,a});
    G[v].push_back({u,a});
    G2[u].push_back({v,b});
    G2[v].push_back({u,b});
  }
  auto d = dijkstra(s,G);
  auto d2 = dijkstra(t,G2);
  vector<ll> ans(n+1,INF);
  for(int i = n-1; i >= 0; --i){
    ans[i]= min(ans[i+1], d[i]+ d2[i]);
  }
  ll tmp = 1e15;
  for(int i = 0; i < n; ++i){
    cout << tmp-ans[i] << endl;
  }
}
