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>
#include <utility>
#define llint long long
#define inf 1e18

using namespace std;

struct edge{
	llint to, cost;
	edge(){}
	edge(llint a, llint b){
		to = a, cost = b;
	}
};

llint n, m, s, t;
vector<edge> G[100005], G2[100005];
llint dist[100005], dist2[100005];
llint ans[100005];

typedef pair<llint, llint> P;

void dijkstra(vector<edge> G[], llint dist[], llint S)
{
	for(int i = 1; i <= n; i++) dist[i] = inf;
	dist[S] = 0;
	
	priority_queue< P, vector<P>, greater<P> > Q;
	Q.push(make_pair(0, S));
	
	llint v, d;
	while(Q.size()){
		d = Q.top().first;
		v = Q.top().second;
		Q.pop();
		if(dist[v] < d) continue;
		for(int i = 0; i < G[v].size(); i++){
			if(dist[G[v][i].to] > d + G[v][i].cost){
				dist[G[v][i].to] = d + G[v][i].cost;
				Q.push( make_pair(dist[G[v][i].to], G[v][i].to) );
			}
		}
	}
}

int main(void)
{
	cin >> n >> m >> s >> t;
	llint u, v, a, b;
	for(int i = 0; i < m; i++){
		cin >> u >> v >> a >> b;
		G[u].push_back(edge(v, a));
		G[v].push_back(edge(u, a));
		G2[u].push_back(edge(v, b));
		G2[v].push_back(edge(u, b));
	}
	
	dijkstra(G, dist, s);
	dijkstra(G2, dist2, t);
	
	llint mn = inf;
	for(int i = n; i >= 1; i--){
		mn = min(dist[i] + dist2[i], mn);
		ans[i] = mn;
	}
	
	for(int i = 1; i <= n; i++){
		cout << 1000000000000000LL - ans[i] << endl;
	}
	return 0;
}