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 ;
const int MAX = 1e5 + 10 ;

vector< vector< pair<int, pair<ll , ll> > > >adj(MAX) ;
long long dist[2][MAX] ;
int n , m , s , t ;

void dijkstra(int src , int st)
{
	for(int i = 1 ; i <= n ; ++i)
		dist[st][i] = 1e18 ;
	dist[st][src] = 0 ;
	priority_queue< pair<ll , int> , vector< pair<ll, int> > , greater< pair<ll , int> > >q ;
	q.push({0 , src}) ;
	dist[st][src] = 0 ;
	while(!q.empty())
	{
		pair<ll , int>p = q.top() ;
		q.pop() ;
		int node = p.second ;
		ll cost = p.first ;
		if(dist[st][node] < cost)
			continue ;
		for(auto &child : adj[node])
		{
			int to = child.first ;
			ll tocost = cost ;
			if(st == 0)
				tocost += child.second.first ;
			else
				tocost += child.second.second ;
			if(tocost < dist[st][to])
			{
				dist[st][to] = tocost ;
				q.push({tocost , to}) ;
			}
		}
	}
	return ;
}
int main()
{
	ios_base::sync_with_stdio(0) ;
	cin.tie(0) ;
	cin>>n>>m>>s>>t ;
	for(int i = 0 ; i < m ; ++i)
	{
		int x , y ;
		long long z1 , z2;
		cin>>x>>y>>z1>>z2 ;
		adj[x].push_back({y , {z1 , z2}}) ;
		adj[y].push_back({x , {z1 , z2}}) ;
	}
	dijkstra(s , 0) ;
	dijkstra(t , 1) ;
	ll cons = 1e15 ;
	ll suff[n+1] ;
	suff[n] = dist[0][n] + dist[1][n] ;
	for(int i = n-1 ; i >= 0 ; --i)
		suff[i] = min(suff[i+1] , dist[0][i] + dist[1][i]) ;
	for(int i = 1 ; i <= n ; ++i)
		cout<<cons-suff[i]<<"\n" ;
	return 0 ;
}