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,m,n) for(int i=m;i<n;i++)

typedef long long ll;
typedef pair<ll,ll> P;

const ll inf=1000000000000000;
 struct edge{ll to,cost;};

void dijkstra(ll s,vector<vector<edge>> &G, vector<ll> &d){
priority_queue<P,vector<P>,greater<P>> que;
d[s]=0;
que.push(P(0,s));
while(!que.empty()){
 P p=que.top(); que.pop();
ll v=p.second;
if(d[v]<p.first) continue;
for(int i=0;i<G[v].size();i++){
 edge e=G[v][i];

if(d[e.to]>d[v]+e.cost){
d[e.to]=d[v]+e.cost;
que.push(P(d[e.to],e.to));
}
}
}
}
  
int main(){
  cin.tie(nullptr);
  ios::sync_with_stdio(false);
  
  ll n,m,s,t;
  cin >> n >> m >> s >> t;
  s--;
  t--;
  vector<vector<edge>> G(n),H(n);
  
  rep(i,0,m){
   ll u,v,a,b;
    cin >> u >> v >> a >> b;
    u--;
    v--;
   G[u].push_back(edge{v,a});
   G[v].push_back(edge{u,a});
   H[u].push_back(edge{v,b});
   H[v].push_back(edge{u,b});
  }
  
  vector<ll> d(n,inf),e(n,inf);
  dijkstra(s,G,d);
  dijkstra(t,H,e);
  

  
  vector<ll> res(n);
    res[n-1]=d[n-1]+e[n-1];
  for(int i=n-2;i>=0;i--) res[i]=min(res[i+1],d[i]+e[i]);
  
  rep(i,0,n) cout << inf-res[i] << endl;
}
