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>
#include <unordered_set>
using namespace std;
#define ll long long
#define rep(i,n) for(int (i)=0;(i)<(n);(i)++)
#define repeat(i,s,n) for(int (i)=s; (i)<(n); (i)++)
#define revrep(i,n) for(int (i)=(n)-1;i>=0; i--)
using WG = map<int,map<int,ll>>; // weighted graph

void dijkstra(int s, WG& g, vector<ll>& dist) {
  vector<bool> visited(dist.size());
  fill(dist.begin(),dist.end(),1e15);
  using P = pair<ll,int>;
  priority_queue<P,vector<P>, greater<P>> pq;
  dist[s]=0;
  //visited[s]=true;
  pq.push(make_pair(0LL,s));
  while(!pq.empty()) {
    int v;
    ll d;
    tie(d,v)=pq.top();
    pq.pop();
    if(visited[v]) continue;
    visited[v]=true;
    for(auto& uw : g[v]) {
      int u;
      ll w;
      tie(u,w)=uw;
      if(visited[u]) continue;
      if(d+w<dist[u]) {
        dist[u]=d+w;
        pq.push(make_pair(dist[u],u));
      }
    }
  }
}

int main() {
  cin.tie(0);
  ios::sync_with_stdio(false);
  cout<<setprecision(std::numeric_limits<float>::max_digits10);
  int n,m,s,t;
  cin>>n>>m>>s>>t;
  vector<ll> YD(n+1);
  vector<ll> SD(n+1);
  WG YG;
  WG SG;
  rep(__i,m) {
    int u,v;
    ll a,b;
    cin>>u>>v>>a>>b;
    YG[u][v]=a;
    YG[v][u]=a;
    SG[u][v]=b;
    SG[v][u]=b;
  }
  dijkstra(s,YG,YD);
  dijkstra(t,SG,SD);
  ll a=0;
  const ll y=1e15;
  vector<ll> ans(n);
  revrep(i,n) {
    a=max(a,y-(YD[i+1]+SD[i+1]));
    ans[i]=a;
//    cout << a << endl;
  }
  rep(i,n) cout<<ans[i]<<endl;
  return 0;
}
