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;
vector<pair<int, ll> > adj[100010], adj2[100010];
int n, m;
void dij(int s, vector<pair<int, ll> > *adj, ll *dis)
{
	for (int i = 1; i <= n; i++) dis[i] = 1e18;
	dis[s] = 0;
	priority_queue<pair<ll, int>, vector<pair<ll, int> >, greater<pair<ll, int> > > pq;
	pq.emplace(0, s);
	while (pq.size())
	{
		int a = pq.top().second;
		ll d = pq.top().first;
		pq.pop();
		if (dis[a] != d) continue;
		for (auto c : adj[a])
		{
			if (d + c.second < dis[c.first])
			{
				dis[c.first] = d + c.second;
				pq.emplace(dis[c.first], c.first);
			}
		}
	}
}
ll dis1[100010], dis2[100010];
ll ans[100010];
int s, t;
int main()
{
	scanf("%d%d%d%d", &n, &m, &s, &t);
	for (int i = 0; i < m; i++)
	{
		int a, b;
		ll c, d;
		scanf("%d%d%lld%lld", &a, &b, &c, &d);
		adj[a].emplace_back(b, c);
		adj[b].emplace_back(a, c);
		adj2[a].emplace_back(b, d);
		adj2[b].emplace_back(a, d);
	}
	dij(s, adj, dis1);
	dij(t, adj2, dis2);
	for (int i = 1; i <= n; i++)
	{
		ans[i] = 1e15 - dis1[i] - dis2[i];
	}
	for (int i = n-1; i >= 1; i--)
	{
		ans[i] = max(ans[i], ans[i+1]);
	}
	for (int i = 1; i <= n; i++) printf("%lld\n", ans[i]);
}