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 <iostream>
#include <bits/stdc++.h>
#define F first
#define S second
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;
ll n,m,s,g,aa,bb,a,b,loc1[100000],loc2[100000],flag[100000];
int main(void){
  cin>>n>>m>>s>>g;
  s--,g--;
  vector<vector<P>>G1(n),G2(n);
  for(int i=0;i<m;i++){
    cin>>aa>>bb>>a>>b;
    aa--,bb--;
    G1[aa].push_back({bb,a});
    G1[bb].push_back({aa,a});
    G2[aa].push_back({bb,b});
    G2[bb].push_back({aa,b});
  }
  for(int i=0;i<n;i++)loc1[i]=(1LL<<60);
  priority_queue<P,vector<P>,greater<P>>pque;
  pque.push({0,s});
  loc1[s]=0;
  while(pque.size()){
    auto p=pque.top();
    pque.pop();
    if(!flag[p.S]){
      flag[p.S]=1;
      for(auto x:G1[p.S]){
        if(loc1[p.S]+x.S<loc1[x.F]){
          loc1[x.F]=loc1[p.S]+x.S;
          pque.push({loc1[x.F],x.F});
        }
      }
    }
  }
  for(int i=0;i<n;i++){loc2[i]=(1LL<<60);flag[i]=0;}
  pque.push({0,g});
  loc2[g]=0;
  while(pque.size()){
    auto p=pque.top();
    pque.pop();
    if(!flag[p.S]){
      flag[p.S]=1;
      for(auto x:G2[p.S]){
        if(loc2[p.S]+x.S<loc2[x.F]){
          loc2[x.F]=loc2[p.S]+x.S;
          pque.push({loc2[x.F],x.F});
        }
      }
    }
  }
  for(int i=0;i<n;i++)loc1[i]+=loc2[i];
  for(int i=n-1;i>=1;i--){
    loc1[i-1]=min(loc1[i-1],loc1[i]);
  }
  for(int i=0;i<n;i++)cout<<1000000000000000LL-loc1[i]<<endl;
}
