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<cstdio>
#include<cstdlib>
#include<algorithm>
#include<iostream>
#include<queue>
#include<vector>
#include <bitset>
#include<math.h>
using namespace std;
#define INF 1100000000000000
#define MOD 1000000007
typedef long long ll;
typedef pair<int,ll> P;

struct edge{
  int to;
  ll cost;
};

int V;
ll S[110000],T[110000];
vector<P> SV[110000],TV[110000];
int d[110000];

void dijkstra(int s,vector<P> G[],ll d[]){
  // greater<P>を指定することでfirstが小さい順に取り出せる
  priority_queue<P,vector<P>,greater<P>> que;
  fill(d,d+100001,INF);
  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<(int)G[v].size();i++){
      P e=G[v][i];
      if(d[e.first]>d[v]+e.second){
        d[e.first]=d[v]+e.second;
        que.push(P(d[e.first],e.first));
      }
    }
  }
}

int main(){
  ll a,b,ans[110000];
  int n,m,s,t,u,v;
  cin>>n>>m>>s>>t;
  ans[n+1]=INF;
  for(int i=0;i<m;i++){
    cin>>u>>v>>a>>b;
    SV[u].push_back(P(v,a));
    SV[v].push_back(P(u,a));

    TV[u].push_back(P(v,b));
    TV[v].push_back(P(u,b));
  }

  dijkstra(s,SV,S);
  dijkstra(t,TV,T);

  for(int i=n;i>0;i--){
    ans[i]=min(ans[i+1],S[i]+T[i]);
    //cout<<S[i]<<" "<<T[i]<<endl;
  }
  for(int i=1;i<=n;i++){
    cout<<1000000000000000-ans[i]<<endl;
  }
}
