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;

vector<pair<int, int>> g[111111];
int a[111111], b[111111];
long c[111111], d[111111];
int main() {
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	for (int i = 1; i <= m; i++) {
		int u, v;
		scanf("%d%d%d%d", &u, &v, a + i, b + i);
		g[u].emplace_back(i, v);
		g[v].emplace_back(i, u);
	}
	fill(c, c + n + 1, 1e18);
	c[s] = 0;
	priority_queue<pair<long, int>, vector<pair<long, int>>, greater<pair<long, int>>> q;
	q.emplace(0, s);
	while (!q.empty()) {
		int u = q.top().second;
		q.pop();
		for (auto v : g[u]) {
			auto l = c[u] + a[v.first];
			if (c[v.second] <= l)
				continue;
			c[v.second] = l;
			q.emplace(c[v.second], v.second);
		}
	}
	fill(d, d + n + 1, 1e18);
	d[t] = 0;
	q.emplace(0, t);
	while (!q.empty()) {
		int u = q.top().second;
		q.pop();
		for (auto v : g[u]) {
			auto l = d[u] + b[v.first];
			if (d[v.second] <= l)
				continue;
			d[v.second] = l;
			q.emplace(d[v.second], v.second);
		}
	}
	vector<long> r;
	long M = 1e18;
	for (int i = n; i; i--)
		r.push_back(M = min(M, c[i] + d[i]));
	for (auto it = r.rbegin(); it != r.rend(); ++it)
		cout << long(1e15) - *it << '\n';
	cout << flush;
	return 0;
}
