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> vl ;
typedef vector<int> vi;
typedef pair<int, int> pi;
typedef pair<ll,ll> pl;

#define rep(i,N) for(int i=0; i < (int)N; i++)
#define all(v) (v).begin(),(v).end()

const ll INF = (ll) 1e18 + 5;

struct edge{
	int to, cost;
	edge(int to, int cost):to(to), cost(cost){}
};

const int MAX_V = 100005;
int n;

vl dijkstra(int s, vector<vector<edge>> & g){
	priority_queue<pi, vector<pi>, greater<pi>> que;
	vl d(n, 1e18);
	d[s] = 0;
	que.push(pi(0, s));
	while(!que.empty()){
		pi p = que.top(); que.pop();
		int v = p.second;
		if (d[v] < p.first) continue;
		for (auto &e : g[v]){
			ll cost = e.cost;
			int u = e.to;
			if (d[u] > d[v] + cost){
				d[u] = d[v] + cost;
				que.emplace(d[u], u);
			}
		}
	}

	return d;
}

int main(){
	int m, s, t; cin >> n >> m >> s >> t;
	s--; t--;
	vector<vector<edge>> g0(n), g1(n);
	rep(i,m){
		int u, v, a, b; cin >> u >> v >> a >> b;
		u--; v--;
		g0[u].emplace_back(v, a);
		g0[v].emplace_back(u, a);
		g1[u].emplace_back(v, b);
		g1[v].emplace_back(u, b);
	}
	vl da = dijkstra(s, g0);
	vl db = dijkstra(t, g1);
	vl d(n, 1e15);
	d[n-1] -= da[n-1] + db[n-1];
	for(int i = n-2; i>=0; i--){
		d[i] = max(d[i+1], d[i] - da[i] - db[i]);
	}
	rep(i,n){
		cout << d[i] << endl;
	}
}