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
// D.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <queue>
#include <map>

using namespace std;

typedef long long LL;
typedef pair<LL, LL> LLLL;
typedef pair<LL, int> LLI;

int main(int argc, char *argv[]) {
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	--s, --t;
	map<int, LLLL> edges[100000];
	for (int i = 0; i < m; ++i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		--u, --v;
		edges[u][v] = LLLL(a, b);
		edges[v][u] = LLLL(a, b);
	}

	priority_queue<LLI, vector<LLI> > queue;

	LL mcy[100000];
	fill(mcy, mcy + 100000, -(1LL << 60));
	mcy[s] = 0;
	queue.push(LLI(0, s));
	while (queue.size()) {
		LLI top = queue.top();
		queue.pop();
		LL x = top.first;
		LL node = top.second;
		for (auto kv : edges[node]) {
			int next = kv.first;
			LL y = x - kv.second.first;
			if (y > mcy[next]) {
				mcy[next] = y;
				queue.push(LLI(y, next));
			}
		}
	}

	LL mcs[100000];
	fill(mcs, mcs + 100000, -(1LL << 60));
	mcs[t] = 0;
	queue.push(LLI(0, t));
	while (queue.size()) {
		LLI top = queue.top();
		queue.pop();
		LL x = top.first;
		LL node = top.second;
		for (auto kv : edges[node]) {
			int next = kv.first;
			LL y = x - kv.second.second;
			if (y > mcs[next]) {
				mcs[next] = y;
				queue.push(LLI(y, next));
			}
		}
	}

	vector<LL> ans(n + 1);
	for (int i = n - 1; i >= 0; --i) {
		ans[i] = max(ans[i + 1], 1000000000000000LL + mcy[i] + mcs[i]);
	}
	for (int i = 0; i < n; ++i) {
		cout << ans[i] << endl;
	}
	return 0;
}
