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;
const long long INF = 1LL<<60;
void calcDist(vector<long long>& dist, const vector<vector<pair<int,int>>>& G, int v0) {
	priority_queue<pair<long long,int>> Q;
	dist[v0] = 0;
	Q.push({0, v0});
	while (!Q.empty()) {
		long long d = -Q.top().first;
		int v = Q.top().second;
		Q.pop();
		if (dist[v] != d) continue;
		for (auto p : G[v]) {
			int vv = p.first;
			long long dd = d + p.second;
			if (dd < dist[vv]) {
				dist[vv] = dd;
				Q.push({-dd, vv});
			}
		}
	}
	//for (auto x : dist) cout << x << " "; cout << endl;
}
int main() {
	int N, M, S, T;
	cin >> N >> M >> S >> T;
	-- S; -- T;
	vector<vector<pair<int,int>>> G1(N), G2(N);
	for (int i = 0; i < M; ++ i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		-- u; -- v;
		G1[u].push_back({v,a});
		G1[v].push_back({u,a});
		G2[u].push_back({v,b});
		G2[v].push_back({u,b});
	}
	vector<long long> dist1(N, INF), dist2(N, INF);
	calcDist(dist1, G1, S);
	calcDist(dist2, G2, T);
	vector<long long> r(N);
	long long rr = INF;
	for (int i = N-1; i >= 0; -- i) {
		//cerr << dist1[i] + dist2[i] << endl;
		rr = min(rr, dist1[i] + dist2[i]);
		r[i] = rr;
	}
	for (auto x : r) cout << (1000000000000000LL-x) << endl;
}
