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 int long long
using P=pair<int,int>;
using PP=pair<int,P>;
signed main(){
  int n,m,s,t;
  cin>>n>>m>>s>>t;
  s--,t--;
  vector<vector<PP>> A(n);
  for(int i=0;i<m;i++){
    int u,v,a,b;
    cin>>u>>v>>a>>b;
    A[u-1].push_back({v-1,{a,b}});
    A[v-1].push_back({u-1,{a,b}});
  }
  vector<int> B(n,1000000000000000000),C(n,1000000000000000000);
  B[s]=0,C[t]=0;
  priority_queue<P,vector<P>,greater<P>> Q;
  Q.push({s,0});
  while(!Q.empty()){
    P R=Q.top();
    Q.pop();
    if(R.second>B[R.first])
      continue;
    for(PP p:A[R.first])
      if(B[p.first]>R.second+p.second.first){
        B[p.first]=R.second+p.second.first;
        Q.push({p.first,R.second+p.second.first});
      }
  }
  Q.push({t,0});
  while(!Q.empty()){
    P R=Q.top();
    Q.pop();
    if(R.second>C[R.first])
      continue;
    for(PP p:A[R.first])
      if(C[p.first]>R.second+p.second.second){
        C[p.first]=R.second+p.second.second;
        Q.push({p.first,R.second+p.second.second});
      }
  }
  vector<int> D(n+1,0);
  for(int i=n-1;i>=0;i--)
    D[i]=max(D[i+1],1000000000000000-(B[i]+C[i]));
  for(int i=0;i<n;i++)
    cout<<D[i]<<endl;
}