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>
#define mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;
vector <ll> a[100010], b1[100010], b2[100010];
ll ans[100010], dis1[100010], dis2[100010];
int main(){
  ll n, m, aa, bb, x, y, s, t;
  while(cin>>n>>m>>s>>t){
    for(int i=0; i<m; ++i){
      cin>>x>>y>>aa>>bb;
      a[x].push_back(y);
      b1[x].push_back(aa);
      b2[x].push_back(bb);

      a[y].push_back(x);
      b1[y].push_back(aa);
      b2[y].push_back(bb);
    }
    for(int i=1; i<=n; ++i){
      dis1[i]=dis2[i]=1e17;
    }
    priority_queue < pair<ll, ll> > q;
    q.push(mk(0, s));
    pair<ll, ll> tp, tmp;
    while(!q.empty()){
      tp=q.top();
      q.pop();
      tp.fs*=-1;
      if(dis1[tp.sc]<=tp.fs)
        continue;
      dis1[tp.sc]=tp.fs;
      for(int i=0; i<a[tp.sc].size(); ++i){
        q.push(mk(-tp.fs-b1[tp.sc][i], a[tp.sc][i]));
      }
    }
    q.push(mk(0, t));
    while(!q.empty()){
      tp=q.top();
      q.pop();
      tp.fs*=-1;
      if(dis2[tp.sc]<=tp.fs)
        continue;
      dis2[tp.sc]=tp.fs;
      for(int i=0; i<a[tp.sc].size(); ++i){
        q.push(mk(-tp.fs-b2[tp.sc][i], a[tp.sc][i]));
      }
    }
    ans[n+1]=1e17;
    for(int i=n; i>=1; --i){
      ans[i]=min(ans[i+1], dis1[i]+dis2[i]);
    }
    ll mx=1e15;
    for(int i=1; i<=n; ++i){
      cout<<mx-ans[i]<<endl;
    }
  }
  return 0;
}


