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;
const ll INF = 10000000000000000LL;
const ll MAX = 1000000000000000LL;
struct edge{ ll to,cost; };
typedef pair<ll,ll> P;

vector<ll> dijkstra(vector< vector< edge > > G, int si){
  vector<ll> d( G.size() , INF );
  d[si]=0;
  priority_queue< P , vector<P> , greater<P> > Q;
  Q.push( P(0,si) );
  while(!Q.empty()){
    P p=Q.top();
    Q.pop();
    ll pos=p.second;
    ll cost=p.first;
    if( d[ pos ] < cost )continue;
    for(int i=0;i<(int)G[pos].size();i++){
      edge e=G[pos][i];
      ll ncost=e.cost+cost;
      if(ncost < d[e.to]){
        d[e.to]=ncost;
        Q.push( P(ncost,e.to));
      }
    }
  }
  return d;
}

int main(){
  vector< vector< edge > > GA;
  vector< vector< edge > > GB;
  
  int n,m,s,t;
  cin>>n>>m>>s>>t;
  GA.resize(n);
  GB.resize(n);
  
  s--,t--;
  for(int i=0;i<m;i++){
    ll u,v,a,b;
    cin>>u>>v>>a>>b;
    u--,v--;
    GB[u].push_back((edge){v,b});
    GB[v].push_back((edge){u,b});

    GA[u].push_back((edge){v,a});
    GA[v].push_back((edge){u,a});
  }

  vector<ll> DA = dijkstra( GA , s );
  vector<ll> DB = dijkstra( GB , t );

  vector<ll> ans(n, INF);
  for(int i=n-1;i>=0;i--){
    ans[i]=DA[i]+DB[i];
    if(i+1 < n)ans[i]=min(ans[i],ans[i+1]);
  }

  for(int i=0;i<n;i++){
    cout<<MAX-ans[i]<<endl;
  }
  return 0;
}
