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;
typedef long long ll;
const ll INF = 1LL << 60;
const ll MOD = 1e9 + 7;
#define rep(i,n) for (int i = 0; i < (n); i++)
#define ALL(v) v.begin(), v.end()
struct edge {
	int to;
	ll weight;
	edge(int to, ll weight = 1) :to(to), weight(weight) {}
};
void dijkstra(int s, vector<vector<edge>> &G, vector<ll> &dist) {
	using P = pair<ll, int>;
	dist.resize(G.size());
	fill(ALL(dist), INF);
	dist[s] = 0;
	priority_queue<P, vector<P>, greater<P>> Q;
	Q.emplace(0, s);
	while (!Q.empty()) {
		P now = Q.top(); Q.pop();
		if (dist[now.second] < now.first) continue;
		for (edge &e : G[now.second]) {
			if (dist[e.to] > now.first + e.weight) {
				dist[e.to] = now.first + e.weight;
				Q.emplace(dist[e.to], e.to);
			}
		}
	}
}
int main() {
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	s--; t--;
	vector<vector<edge>> G1(n), G2(n);
	rep(i, m) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		u--; v--;
		G1[u].emplace_back(v, a);
		G1[v].emplace_back(u, a);
		G2[u].emplace_back(v, b);
		G2[v].emplace_back(u, b);
	}
	vector<ll> dist1(n), dist2(n);
	dijkstra(s, G1, dist1);
	dijkstra(t, G2, dist2);
	vector<ll> ans(n);
	for (int i = n - 1; i >= 0; i--) {
		ll snuke = 1e15;
		snuke -= dist1[i];
		snuke -= dist2[i];
		ans[i] = snuke;
		if (i + 1 < n) ans[i] = max(ans[i], ans[i + 1]);
	}
	rep(i, n) {
		cout << ans[i] << endl;
	}
}