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 vector<ll> vec;
typedef vector<vec> mat;
typedef pair<ll,ll> pll;
const ll mod=1e9+7;
//const ll mod=998244353;
const ll inf=5e18;

int main() {
  ll n,m,s,g;
  cin >> n >> m >> s >> g;
  s--;g--;
  vector<map<ll,ll>> Ea(n),Eb(n);
  for(ll i=0;i<m;i++) {
    ll f,t,a,b;
    cin >> f >> t >> a >> b;
    f--;t--;
    Ea[f][t]=Ea[t][f]=a;
    Eb[f][t]=Eb[t][f]=b;
  }
  vec da(n,inf),db(n,inf);
  priority_queue<pll,vector<pll>,greater<pll>> q;
  q.push(pll(0,s));
  while(q.size()) {
    ll f=q.top().second;
    ll d=q.top().first;
    q.pop();
    if(da[f]!=inf) continue;
    da[f]=d;
    for(auto p:Ea[f]) {
      ll t=p.first;
      ll nd=d+p.second;
      if(da[t]!=inf) continue;
      q.push(pll(nd,t));
    }
  }
  q.push(pll(0,g));
  while(q.size()) {
    ll f=q.top().second;
    ll d=q.top().first;
    q.pop();
    if(db[f]!=inf) continue;
    db[f]=d;
    for(auto p:Eb[f]) {
      ll t=p.first;
      ll nd=d+p.second;
      if(db[t]!=inf) continue;
      q.push(pll(nd,t));
    }
  }
  vec ans(n);
  ans[n-1]=da[n-1]+db[n-1];
  for(ll i=n-2;i>-1;i--) {
    ans[i]=min(ans[i+1],da[i]+db[i]);
  }
  for(ll i=0;i<n;i++) {
    cout << (ll)1e15-ans[i] << endl;
  }
}