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 <iostream>
#include <vector>
#include <queue>
using namespace std;

#define int long long

const int INF = 0x3f3f3f3f3f3f3f3f;
const int INITIAL = 1000000000000000LL;

struct edge {
	int v, w;
	bool operator>(const edge &other) const {
		return w > other.w;
	}
};

void dijkstra(int source, vector< vector<edge> > &adj, vector<int> &dist) {
	priority_queue<edge, vector<edge>, greater<edge> > pq;
	pq.push({source, 0});
	dist[source] = 0;
	while (pq.size()) {
		int u = pq.top().v;
		int w = pq.top().w;
		pq.pop();
		if (w > dist[u]) 
			continue;

		for (edge e: adj[u]) {
			if (w + e.w < dist[e.v]) {
				dist[e.v] = w+e.w;
				pq.push({e.v, w+e.w});
			}
		}
	}
}

signed main() {
	int N, M, S, T;
	cin >> N >> M >> S >> T;
	--S, --T;
	vector< vector<edge> > adj(N), inv(N);
	for (int i = 0; i < M; ++i) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		--u, --v;
		adj[u].push_back({v, a});
		adj[v].push_back({u, a});
		inv[u].push_back({v, b});
		inv[v].push_back({u, b});
	}
	vector<int> distS(N, INF), distT(N, INF);
	dijkstra(S, adj, distS);
	dijkstra(T, inv, distT);
	int mn = INF;
	vector<int> outp;
	for (int i = N-1; i >= 0; --i) {
		mn = min(mn, distS[i]+distT[i]);
		outp.push_back(INITIAL-mn);
	}
	for (int i = N-1; i >= 0; --i) {
		cout << outp[i] << endl;
	}
}
