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 i64 = long long;
#define rep(i,s,e) for(int (i) = (s);(i) <= (e);(i)++)
#define all(x) x.begin(),x.end()

int N,M;

int s,t;

struct edge{
  int to;
  i64 idx;
};

vector<edge> G[101010];

vector<i64> A(101010);
vector<i64> B(101010);

vector<i64> dijkstra(int s,vector<i64>& cost){
  vector<i64> dist(101010,1e18);
  dist[s] = 0;
  using P = pair<i64,i64>;
  priority_queue<P,vector<P>,greater<P>> que;
  que.push({dist[s],s});

  while(!que.empty()){
    int v = que.top().second;
    i64 d = que.top().first;
    que.pop();
    if(dist[v] < d) continue;

    for(auto & e : G[v]){
      if(dist[e.to] > dist[v] + cost[e.idx]){
        dist[e.to] = dist[v] + cost[e.idx];
        que.push({dist[e.to],e.to});
      }
    }
  }
  return dist;
}
int main(){
  cin >> N >> M >> s >> t;
  rep(i,0,M - 1){
    int u,v;
    cin >> u >> v >> A[i] >> B[i];
    G[u].push_back({v,i});
    G[v].push_back({u,i});
  }

  auto money = dijkstra(s,A);
  auto snuuk = dijkstra(t,B);

  i64 ans = 0;
  vector<i64> re;
  i64 base = 1e15;

  for(int i = N;i >= 1;i--){
    re.push_back(ans = max(ans , base - money[i] - snuuk[i]));
  }

  reverse(all(re));

  for(auto x : re){
    cout << x << endl;
  }
}
