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 int MAXN = 100100;
int A[MAXN];

struct edge {
	int v;
	long long w;
	edge() {}
	edge(int v, long long w) : v(v), w(w) {};
};

vector<long long> dijkstra(int n, vector<vector<edge> >& G, int s) {
	vector<long long> d(n, 1e18);
	d[s] = 0;
	priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> que;
	que.push(make_pair((long long)0, s));
	while (!que.empty()) {
		auto p = que.top();
		que.pop();
		int u = p.second;
		long long dist = p.first;
		if (dist > d[u]) continue;
		for (edge e : G[u]) {
			if (d[e.v] > d[u] + e.w) {
				d[e.v] = d[u] + e.w;
				que.push(make_pair(d[e.v], e.v));
			}
		}
	}
	return d;
}

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);
	int N, M, S, T;
	cin >> N >> M >> S >> T;
	S--;
	T--;
	vector<vector<edge> > G(N), rG(N);
	for (int i = 0; i < M; i++) {
		int a, b, c, d;
		cin >> a >> b >> c >> d;
		a--; b--;
		G[a].emplace_back(b, c);
		G[b].emplace_back(a, c);
		rG[a].emplace_back(b, d);
		rG[b].emplace_back(a, d);
	}
	auto d = dijkstra(N, G, S);
	auto rd = dijkstra(N, rG, T);
	long long ans = 0;
	vector<long long>Ans(N);
	for (int i = 0; i < N; i++) {
		long long num = d[N - 1 - i] + rd[N - 1 - i];
		if (0 == i) {
			Ans[N - 1 - i] = num;
		}
		else {
			Ans[N - 1 - i] = min(num, Ans[N - i]);
		}
	}
	for (int i = 0; i < N; i++) {
		long long ans = 1e15;
		ans -= Ans[i];
		cout << ans << endl;
	}
	return 0;
}