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;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;
#define rep(i,n) for(ll i = 0;i < (ll)n;i++)
#define ALL(x) (x).begin(),(x).end()
#define MOD 1000000007

struct edge{
  ll to,a;
};

int main(){
  
  int n,m,s,t;
  scanf("%d %d %d %d",&n,&m,&s,&t);
  s--,t--;
  vector<vector<edge>> v(n),w(n);
  rep(i,m){
    ll u,u_,a,b;scanf("%lld %lld %lld %lld",&u,&u_,&a,&b);
    u--,u_--;
    v[u].push_back({u_,a});
    v[u_].push_back({u,a});
    w[u].push_back({u_,b});
    w[u_].push_back({u,b});
  }
  vector<ll> yen(n,1e18);
  vector<ll> snuuk(n,1e18);
  auto dijkstra = [](vector<vector<edge>> &v,vector<ll> &d,int s){
    priority_queue<pair<ll,ll>,vector<pair<ll,ll>>,greater<pair<ll,ll>>> que;
    que.push(make_pair(0,s));
    d[s] = 0;
    while(!que.empty()){
      pair<ll,ll> ov = que.top();que.pop();
      if(ov.first > d[ov.second])continue;
      for(auto nv : v[ov.second]){
        if(d[nv.to] <= d[ov.second]+nv.a)continue;
        d[nv.to] = d[ov.second]+nv.a;
        que.push(make_pair(d[nv.to],nv.to));
      }
    }
  };
  dijkstra(v,yen,s);
  dijkstra(w,snuuk,t);
  multiset<ll> st;
  rep(i,n)st.insert(yen[i]+snuuk[i]);
  rep(i,n){
    cout << (ll)1e15-(*st.begin()) << endl;
    st.erase(st.find(yen[i]+snuuk[i]));
  }


  return 0;
}