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;
int n, m, s, t;
long long init = 1e15;
vector<long long> dijkstra(vector<vector<pair<int, long long>>> graph, int start){
  vector<long long> ret(n,1e18);
  priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> que;
  que.push({0, start});
  ret[start] = 0;
  while(!que.empty()){
    auto p = que.top();
    que.pop();
    for(auto to: graph[p.second]){
      if(ret[to.first] > p.first + to.second){
        ret[to.first] = p.first + to.second;
        que.push({ret[to.first], to.first});
      }
    }
  }
  return ret;
}
int main(){
  cin >> n >> m >> s >> t;
  s--;
  t--;
  vector<vector<pair<int, long long>>> yen(n), snuuk(n);
  for(int i=0; i<m; i++){
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--;
    v--;
    yen[u].push_back({v, a});
    yen[v].push_back({u, a});
    snuuk[u].push_back({v, b});
    snuuk[v].push_back({u, b});
  }
  auto ycost = dijkstra(yen, s);
  auto scost = dijkstra(snuuk, t);
  vector<long long> cost(n);
  for(int i=0; i<n; i++) cost[i] = ycost[i] + scost[i];
  vector<long long> ans(n+1,0);
  for(int i=n-1; i>-1; i--)  ans[i] = max(ans[i+1], init - cost[i]);
  for(int i=0; i<n; i++) cout << ans[i] << endl;
}