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 N 1000000000000000
#define inf 10000000000000000
using namespace std;

typedef pair<int,int>P;
struct edge{
  int req;
  long long cost[2];
};
vector<edge> data[100005];
long long d[2][100005];
long long ans[100005];
long long n,m,t[2];
void dik(int check);

int main(){
  long long i,j,k;
  int a;
  cin >> n >> m >> t[0] >> t[1];
  for(i=0;i<m;++i){
    edge x;
    cin >> a >>x.req >> x.cost[0] >> x.cost[1];
    data[a].push_back(x);
    swap(a,x.req);
    data[a].push_back(x);
  }
  dik(0);
  dik(1);
  //cout << n <<" "<< d[1][1];
  /*for(j=0;j<2;++j){
    for(i=1;i<=n;++i)cout << d[j][i] <<" ";
    cout << endl;
    }*/
  ans[n+1]=inf;
  for(j=n;j>=1;--j)ans[j]=min(ans[j+1],d[0][j]+d[1][j]);
  for(i=1;i<=n;++i)cout << N-ans[i] << endl;
  return 0;
}

void dik(int check){
  priority_queue< P, vector<P> , greater<P> > qu;
  fill(d[check],d[check]+n+1,inf);
  d[check][t[check]]=0;
  qu.push(P(0,t[check]));
  while(!qu.empty()){
    P p = qu.top();
    qu.pop();
    int  v = p.second;
    if(d[check][v]<p.first)continue;
    for(int i=0;i<data[v].size();++i){
      edge e = data[v][i];
      if(d[check][e.req]>d[check][v]+e.cost[check]){
	d[check][e.req]=d[check][v]+e.cost[check];
	qu.push(P(d[check][e.req],e.req));
      }
    }
    
  }
}
