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 pair<int,int> P;
typedef long long ll;
typedef long double ld;
const int inf=1e9+7;
const ll longinf=1LL<<60;
#define REP(i,m,n) for(int i=(int)(m) ; i < (int) (n) ; ++i )
#define rep(i,n) REP(i,0,n)
#define F first
#define S second

const int mx=100010;
const ll mod=1e9+7;

typedef vector<vector<pair<ll,int>>> wgraph;
typedef pair<ll,int> pli;

vector<ll> bfs(wgraph v,int x){
  priority_queue<pli,vector<pli>,greater<pli>> q;
  vector<ll> dist(v.size(),longinf);
  dist[x]=0;
  q.push({0LL,x});
  
  while(q.size()){
    pli p=q.top(); q.pop();
    ll d=p.F, now=p.S;
    if(dist[now]<d){ continue; }
    for(auto to:v[now]){
      if(dist[to.S]>d+to.F){
        dist[to.S] = d+to.F;
        q.push({dist[to.S], to.S});
      }
    }
  }
  return dist;
}

int main(){
  int n,m,s,t;
  cin >> n >> m >> s >> t;
  s--; t--;
  wgraph v1(n),v2(n);
  rep(i,m){
    int x,y,z1,z2;
    cin >> x >> y >> z1 >> z2; x--; y--;
    v1[x].push_back({z1,y});
    v1[y].push_back({z1,x});
    v2[x].push_back({z2,y});
    v2[y].push_back({z2,x});
  }
  vector<ll> dist1=bfs(v1,s);
  vector<ll> dist2=bfs(v2,t);
  vector<ll> ans(n);
  rep(i,n){ ans[i] = 1e15-dist1[i]-dist2[i]; } //i:exchange point
  rep(i,n-1){ ans[n-i-2]=max(ans[n-i-2],ans[n-i-1]); }//ans[n]
  rep(i,n){ cout << ans[i] <<endl; }
  return 0;
}
