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;
#define rep(i,n) for(int i=0;i<(n);i++)
#define all(a) (a).begin(),(a).end()
#define ll long long
#define P pair<int,int>
int N;
struct edge{int to,cost;};
vector<edge> G[100010];
ll D[100010];
void dijkstra(int s){
  priority_queue<P,vector<P>,greater<P>> Q;
  rep(i,N) D[i]=1000000000000000000;
  D[s]=0;
  Q.push(P(0,s));
  while(!Q.empty()){
    P p=Q.top();
    Q.pop();
    int x=p.second;
    if(D[x]<p.first) continue;
    rep(i,G[x].size()){
      edge e=G[x][i];
      if(D[e.to]>D[x]+e.cost){
        D[e.to]=D[x]+e.cost;
        Q.push(P(D[e.to],e.to));
      }
    }
  }
}
int main(){
  int M,S,T;
  cin>>N>>M>>S>>T;
  int U[M],V[M],A[M],B[M];
  rep(i,M) cin>>U[i]>>V[i]>>A[i]>>B[i];
  rep(i,M){
    edge e1={V[i]-1,A[i]};
    edge e2={U[i]-1,A[i]};
    G[U[i]-1].push_back(e1);
    G[V[i]-1].push_back(e2);
  }
  dijkstra(S-1);
  ll X[N];
  rep(i,N) X[i]=D[i];
  rep(i,N) G[i].clear();
  rep(i,M){
    edge e1={V[i]-1,B[i]};
    edge e2={U[i]-1,B[i]};
    G[U[i]-1].push_back(e1);
    G[V[i]-1].push_back(e2);
  }
  dijkstra(T-1);
  ll ans[N];
  ans[N-1]=1000000000000000-X[N-1]-D[N-1];
  rep(i,N-1) ans[N-2-i]=max(ans[N-1-i],1000000000000000-X[N-2-i]-D[N-2-i]);
  rep(i,N) cout<<ans[i]<<endl;
}