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<algorithm>
#include<vector>
#include<queue>
using namespace std;
int n,m,s,t;
#include<vector>
#include<queue>
#include<limits>
template<typename T>
vector<T>dijkstra(int s,const vector<vector<pair<int,T> > >&G,T INF=numeric_limits<T>::max())
{
	int n=G.size();
	vector<T>d(n,INF);
	vector<int>parent(n,-1);
	priority_queue<pair<T,int>,vector<pair<T,int> >,greater<pair<T,int> > >P;
	d[s]=0;
	P.push(make_pair(d[s],s));
	while(!P.empty())
	{
		pair<T,int>p=P.top();P.pop();
		int v=p.second;
		if(d[v]<p.first)continue;
		for(const pair<int,T>&e:G[v])
		{
			int u=e.first;
			T cost=d[v]+e.second;
			if(d[u]>cost)
			{
				d[u]=cost;
				parent[u]=v;
				P.push(make_pair(d[u],u));
			}
		}
	}
	return d;
}
int main()
{
	cin>>n>>m>>s>>t;
	s--,t--;
	vector<vector<pair<int,long> > >G(n),H(n);
	for(int i=0;i<m;i++)
	{
		int u,v,a,b;cin>>u>>v>>a>>b;u--,v--;
		G[u].push_back(make_pair(v,a));
		G[v].push_back(make_pair(u,a));
		H[u].push_back(make_pair(v,b));
		H[v].push_back(make_pair(u,b));
	}
	vector<long>ds=dijkstra(s,G);
	vector<long>dt=dijkstra(t,H);
	long ans=1e18;
	vector<long>s;
	for(int i=n;i--;)
	{
		ans=min(ans,ds[i]+dt[i]);
		s.push_back(1e15-ans);
	}
	for(int i=n;i--;)cout<<s[i]<<endl;
}
