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>
typedef long long i64;
using std::cout;
using std::endl;
using std::cin;

int n;
std::vector<std::vector<std::pair<int, std::pair<i64, i64>>>> g;

std::vector<i64> solve(int s, bool f) {
	typedef std::pair<i64, int> P;
	std::priority_queue<P, std::vector<P>, std::greater<P>> qu;
	std::vector<i64> min_cost(n, 1LL << 60); min_cost[s] = 0; qu.push({0, s});
	while(!qu.empty()) {
		auto p = qu.top(); qu.pop();
		
		for(auto e : g[p.second]) {
			i64 cost = p.first;
			if(f) cost += e.second.first;
			else cost += e.second.second;
			if(cost >= min_cost[e.first]) continue;
			min_cost[e.first] = cost;
			qu.push({cost, e.first});
		}
	}
	
	return min_cost;
}

int main() {
	int m, s, t; cin >> n >> m >> s >> t; s--; t--; g.resize(n);
	for(int i = 0; i < m; i++) {
		i64 a, b, c, d; cin >> a >> b >> c >> d;
		
		g[a - 1].push_back({b - 1, {c, d}});
		g[b - 1].push_back({a - 1, {c, d}});
	}
	auto A = solve(s, true), B = solve(t, false);
	
	i64 mi = 1LL << 60;
	std::vector<i64> ans;
	for(int i = n - 1; i >= 0; i--) {
		mi = std::min(mi, A[i] + B[i]);
		ans.push_back(i64(1e15) - mi);
	}
	reverse(begin(ans), end(ans));
	for(auto v : ans) cout << v << endl;
	return 0;
}
