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 <tuple>
#include <vector>
#include <queue>

using namespace std;

typedef long long ll;
typedef tuple<ll, ll> ii;
typedef vector<bool> vb;
typedef vector<ll> vi;
typedef vector<ii> vii;
typedef vector<vii> vvii;
typedef priority_queue<ii, vii, greater<ii>> pq;

ll n, m, s, t, u, v, a, b;

void bfs(vvii &g, vi& c, int s) {
	vb v(n);
	pq q;
	q.push(make_tuple(0, s));
	while (!q.empty()) {
		ll k, i;
		tie(k, i) = q.top();
		q.pop();
		if (v[i]) continue;
		v[i] = true;
		c[i] = k;
		for (auto &z:g[i]) {
			ll k2, i2;
			tie(k2, i2) = z;
			if (v[i2]) continue;
			q.push(make_tuple(k + k2, i2));
		}
	}
}

int main() {
	cin >> n >> m >> s >> t;
	s--; t--;
	vvii ga(n), gb(n);
	for (int i = 0; i < m; i++) {
		cin >> u >> v >> a >> b;
		u--; v--;
		ga[u].push_back(make_tuple(a, v));
		ga[v].push_back(make_tuple(a, u));
		gb[u].push_back(make_tuple(b, v));
		gb[v].push_back(make_tuple(b, u));
	}

	vi ca(n), cb(n), c(n);
	bfs(ga, ca, s);
	bfs(gb, cb, t);
	c[n - 1] = ca[n - 1] + cb[n - 1];
	for (int i = n - 2; i >= 0; i--) c[i] = min(c[i + 1], ca[i] + cb[i]);
	for (int i = 0; i < n; i++) cout << 1000000000000000LL - c[i] << endl;
}
