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> pii;

#define fi first
#define se second
#define mp make_pair
#define fastIO ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());

const int N = (int)1e5 + 9;
const ll inf = (ll)1e15;
vector<pii> T[2][N];

ll d[2][N];
ll ans[N];

int n;
void dij(int node, int q){
  for(int i = 1; i <= n; i ++ ){
    d[q][i] = inf;
  }
  d[q][node] = 0;
  priority_queue<pii,vector<pii>, greater<pii>> ff;
  ff.push(mp(0,node));
  ll dis;
  while(!ff.empty()){
    node = ff.top().se;
    dis = ff.top().fi;
    ff.pop();
    if(dis != d[q][node]) continue;
    for(auto x : T[q][node]){
      if(d[q][node] + x.se < d[q][x.fi]){
        d[q][x.fi] = d[q][node] + x.se;
        ff.push(mp(d[q][x.fi], x.fi));
      }
    }
  }
}

int main(){
  fastIO;
  int m, s, t;
  cin >> n >> m >> s >> t;
  int u, v;
  ll a, b;
  for(int i = 0 ; i < m ; i ++ ){
    cin >> u >> v >> a >> b;
    T[0][u].push_back(mp(v,a));
    T[0][v].push_back(mp(u,a));
    T[1][u].push_back(mp(v,b));
    T[1][v].push_back(mp(u,b));
  }
  dij(s,0);
  dij(t,1);
  for(int i = n; i >= 1; i -- ){
    ans[i] = inf - d[0][i] - d[1][i];
    ans[i] = max(ans[i], ans[i + 1]);
  }
  for(int i = 1; i <= n; i ++ ){
    cout << ans[i] << "\n";
  }
  return 0;
}