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 long long ll;
typedef pair<ll,ll> P;
struct edge{
  ll to,cost;
};
#define MAXN 100010
#define MAXM 100010
#define INF 100000000000000000
ll n,m,s,t;
ll u[MAXM],v[MAXM],a[MAXM],b[MAXM];
vector<edge> Ga[MAXN],Gb[MAXN];
ll da[MAXN],db[MAXN];
int main(){
  cin>>n>>m>>s>>t;
  for(int i=0;i<m;i++){
    cin>>u[i]>>v[i]>>a[i]>>b[i];
    Ga[u[i]].push_back({v[i],a[i]});
    Ga[v[i]].push_back({u[i],a[i]});
    Gb[u[i]].push_back({v[i],b[i]});
    Gb[v[i]].push_back({u[i],b[i]});
  }
  priority_queue<P,vector<P>,greater<P>> qa,qb;
  fill(da,da+MAXN,INF);
  fill(db,db+MAXN,INF);
  da[s]=0;
  db[t]=0;
  qa.push(P(0,s));
  qb.push(P(0,t));
  while(!qa.empty()){
    P p=qa.top();
    qa.pop();
    ll w=p.second;
    if(da[w]<p.first) continue;
    for(int i=0;i<Ga[w].size();i++){
      edge e=Ga[w][i];
      if(da[e.to]>da[w]+e.cost){
        da[e.to]=da[w]+e.cost;
        qa.push(P(da[e.to],e.to));
      }
    }
  }
  while(!qb.empty()){
    P p=qb.top();
    qb.pop();
    ll w=p.second;
    if(db[w]<p.first) continue;
    for(int i=0;i<Gb[w].size();i++){
      edge e=Gb[w][i];
      if(db[e.to]>db[w]+e.cost){
        db[e.to]=db[w]+e.cost;
        qb.push(P(db[e.to],e.to));
      }
    }
  }
  ll d[MAXN];
  d[n]=da[n]+db[n];
  for(int i=n-1;i>0;i--){
    d[i]=min(d[i+1],da[i]+db[i]);
  }
  for(int i=1;i<=n;i++){
    cout<<1000000000000000-d[i]<<endl;
  }
}