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 unsigned long long ull;
typedef long long ll;
typedef pair<ll, int> pr;

#define MAX 100000
#define INF (1ll<<60)

struct edge{
	int to;
	ll cost;
};

int n, m;
vector<edge> yen[MAX+1], snk[MAX+1];
ll cost_y[MAX+1], cost_s[MAX+1];

void dijkstra(int s, vector<edge> g[], ll d[]){
	priority_queue<pr, vector<pr>, greater<pr>> que;
	fill(d, d+n+1, INF);
	d[s]=0;
	que.push(pr(0, s));

	while(!que.empty()){
		pr p=que.top();
		que.pop();
		int v=p.second;
		if(d[v]<p.first) continue;
		for(int i=0; i<g[v].size(); i++){
			edge e=g[v][i];
			if(d[e.to]>d[v]+e.cost){
				d[e.to]=d[v]+e.cost;
				que.push(pr(d[e.to], e.to));
			}
		}
	}
	return;
}

int main(){
	int s, t;
	cin >> n >> m >> s >> t;
	for(int i=0; i<m; i++){
		int u, v;
		ll a, b;
		cin >> u >> v >> a >> b;
		yen[u].push_back(edge{v, a});
		yen[v].push_back(edge{u, a});
		snk[u].push_back(edge{v, b});
		snk[v].push_back(edge{u, b});
	}

	dijkstra(s, yen, cost_y);
	dijkstra(t, snk, cost_s);

	vector<ll> ans(n+1);
	ll m_cost=INF;
	for(int i=n; i>0; i--){
		m_cost=min(m_cost, cost_y[i]+cost_s[i]);
		ans[i]=m_cost;
	}
	for(int i=1; i<=n; i++) cout << 1000000000000000-ans[i] << endl;
	return 0;
}