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 N = 100010;
const long long INF = 1e18;
typedef pair<long long, int> li;

int n, m, s, t;
vector < pair<int,int> > A[N], B[N];
long long dis[N][2];
long long res[N];

void dijkstra(int start) {
	int k = (start == s);
	for (int i = 1; i <= n; ++i) dis[i][k] = INF;
	dis[start][k] = 0;
	
	priority_queue < li, vector <li>, greater <li> > pq;
	pq.push(make_pair(0, start));

	while(!pq.empty()) {
		li top = pq.top(); pq.pop();
		int u = top.second;
		long long du = top.first;

		if (du != dis[u][k]) continue;

		for (auto &e : ((start == s) ? A[u] : B[u])) {
			int cost = e.first, v = e.second;
			if (dis[v][k] > du + cost) {
				pq.push(make_pair(dis[v][k] = du + cost, v));
			}
		}
	}
}

int main() {
	ios_base::sync_with_stdio(false);
	cin.tie(0);

	cin >> n >> m >> s >> t;
	while(m--) {
		int u, v, x, y;
		cin >> u >> v >> x >> y;
		A[u].push_back(make_pair(x, v));
		A[v].push_back(make_pair(x, u));
		B[u].push_back(make_pair(y, v));
		B[v].push_back(make_pair(y, u));
	}

	dijkstra(s); // computed in yen
	dijkstra(t); // computed in snuuk

	res[n] = INF;
	for (int i = n; i >= 1; --i) {
		res[i - 1] = min(res[i], dis[i][1] + dis[i][0]);
	}

	for (int i = 0; i <= n - 1; ++i) {
		long long cur = res[i];
		printf("%lld\n", (long long)1e15 - cur);
	}
}