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;
typedef pair<ll,ll> pll;
typedef pair<ll,pll> plp;
int main(){
  ll inf=1000000000000000000;
  ll n,m,s,t;cin>>n>>m>>s>>t;
  ll u[m],v[m],a[m],b[m];
  vector<plp> h[n+1];
  for(ll i=0;i<m;i++){
    cin>>u[i]>>v[i]>>a[i]>>b[i];
    h[u[i]].push_back(make_pair(v[i],make_pair(a[i],b[i])));
    h[v[i]].push_back(make_pair(u[i],make_pair(a[i],b[i])));
  }
  ll ds[n+1],dt[n+1];
  fill(ds,ds+n+1,inf);
  fill(dt,dt+n+1,inf);
  priority_queue<pll,vector<pll>,greater<pll>>qs,qt;
  qs.push(make_pair((ll)0,s));
  ds[s]=0;
  while(qs.size()>0){
    pll p=qs.top();qs.pop();
    ll pf=p.first,ps=p.second;
    for(ll i=0;i<h[ps].size();i++){
      plp hi=h[ps][i];
      ll ht=hi.first,he=hi.second.first;
      if(ds[ht]>pf+he){
        ds[ht]=pf+he;
        qs.push(make_pair(ds[ht],ht));
      }
    }
  }
  qt.push(make_pair((ll)0,t));
  dt[t]=0;
  while(qt.size()>0){
    pll p=qt.top();qt.pop();
    ll pf=p.first,ps=p.second;
    for(ll i=0;i<h[ps].size();i++){
      plp hi=h[ps][i];
      ll ht=hi.first,hs=hi.second.second;
      if(dt[ht]>pf+hs){
        dt[ht]=pf+hs;
        qt.push(make_pair(dt[ht],ht));
      }
    }
  }
  vector<ll>vv(n+1);
  ll mi=inf;
  for(ll i=n;i>0;i--){
    mi=min(mi,ds[i]+dt[i]);
    vv[i]=mi;
  }
  for(ll i=1;i<=n;i++)cout<<1000000000000000-vv[i]<<endl;
}