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<iostream>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#define INF (1 << 30) - 1
#define LINF 1LL << 60
const int MOD = 1000000007;
using ll = long long;
using P = pair<int, ll>;

struct Edge{
	int to;
	ll a, b;
	Edge(int to, ll a, ll b) : to(to), a(a), b(b){}
};

vector<vector<Edge>> g;
int n, m, s, t;

vector<ll> dijkstra(int start, int flg){
	vector<ll> dist(n, LINF);
  priority_queue<P, vector<P>, greater<P>> que;
  dist[start] = 0;
  que.push({dist[start], start});

  while(!que.empty()){
  	auto v = que.top();
  	que.pop();
    int from = v.second;

  	for(auto nv : g[from]){
  		int cost = (flg ? nv.a : nv.b);
  		if(dist[from] + cost < dist[nv.to]){
  			dist[nv.to] = dist[from] + cost;
        que.push({dist[nv.to], nv.to});
  		}
  	}
  }
  return dist;
}

int main(){
  cin >> n >> m >> s >> t;
  --s;
  --t;
  g.resize(n);
  for(int i = 0; i < m; ++i){
  	int u, v;
  	ll a, b;
  	cin >> u >> v >> a >> b;
  	--u;
  	--v;
  	g[u].push_back(Edge(v, a, b));
  	g[v].push_back(Edge(u, a, b));
  }

  vector<ll> dist_s = dijkstra(s, 1);
  vector<ll> dist_t = dijkstra(t, 0);
  vector<ll> ans(n + 1, LINF); // 
  for(int i = n - 1; i >= 0; --i){
  	ans[i] = min(ans[i + 1], dist_s[i] + dist_t[i]);
  }

  ll p_money = 1e15;
  for(int i = 0; i < n; ++i){
  	cout << p_money - ans[i] << endl;
  }
  return 0;
}
