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"
#define in std::cin
#define out std::cout
#define rep(i,N) for(int i=0;i<N;++i)
typedef long long int LL;
typedef std::pair<LL, LL> P;
struct edge { LL to, cost; };

const LL inf = 1123456789012345;
LL n, m, s, t;
std::vector<std::vector<edge>>G;

std::vector<LL> dijkstra(LL s)
{
	std::vector<LL>dist(2 * n + 1, inf);
	std::priority_queue<P, std::vector<P>, std::greater<P>>que;
	dist[s] = 0;
	que.push(P(0, s));
	while (!que.empty())
	{
		P p = que.top(); que.pop();
		LL v = p.second;
		if (dist[v] < p.first) { continue; }
		rep(i, G[v].size())
		{
			edge e = G[v][i];
			if (dist[e.to] > dist[v] + e.cost)
			{
				dist[e.to] = dist[v] + e.cost;
				que.push(P(dist[e.to], e.to));
			}
		}
	}
	return dist;
}

int main()
{
	in >> n >> m >> s >> t;
	std::vector<LL>u(m), v(m), a(m), b(m);
	std::priority_queue<P, std::vector<P>, std::greater<P>>res;
	G.resize(2 * n + 1);
	rep(i, m)
	{
		in >> u[i] >> v[i] >> a[i] >> b[i];
		G[u[i]].push_back({ v[i],a[i] });
		G[v[i]].push_back({ u[i],a[i] });
		G[u[i] + n].push_back({ v[i] + n,b[i] });
		G[v[i] + n].push_back({ u[i] + n,b[i] });
	}
	auto yen = dijkstra(s);
	auto snk = dijkstra(t + n);
	rep(i, n) res.push(std::make_pair(yen[i + 1] + snk[i + 1 + n], i));
	rep(i, n)
	{
		while (res.top().second < i) res.pop();
		out << 1000000000000000 - res.top().first << std::endl;
	}
	return 0;
}
