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>

#define rep(i,n) for(int i=0;i<(n);i++)

using namespace std;
using lint=long long;

template<class T> struct edge{
	int to;
	T wt;
	edge(int to,const T& wt):to(to),wt(wt){}
};
template<class T> using weighted_graph=vector<vector<edge<T>>>;

template<class T>
void add_undirected_edge(weighted_graph<T>& G,int u,int v,const T& wt){
	G[u].emplace_back(v,wt);
	G[v].emplace_back(u,wt);
}

template<class T>
vector<T> Dijkstra(const weighted_graph<T>& G,int s){
	constexpr T INF=numeric_limits<T>::max();
	int n=G.size();
	vector<T> d(n,INF); d[s]=0;
	priority_queue<pair<T,int>> Q; Q.emplace(0,s);
	while(!Q.empty()){
		T d0=-Q.top().first;
		int u=Q.top().second; Q.pop();
		if(d0>d[u]) continue;
		for(const auto& e:G[u]){
			int v=e.to;
			if(d[v]>d[u]+e.wt) d[v]=d[u]+e.wt, Q.emplace(-d[v],v);
		}
	}
	return d;
}

const long long INF=1LL<<61;

int main(){
	int n,m,s,t; scanf("%d%d%d%d",&n,&m,&s,&t); s--; t--;
	weighted_graph<lint> G1(n),G2(n);
	rep(i,m){
		int u,v;
		lint c1,c2; scanf("%d%d%lld%lld",&u,&v,&c1,&c2); u--; v--;
		add_undirected_edge(G1,u,v,c1);
		add_undirected_edge(G2,u,v,c2);
	}

	auto d1=Dijkstra(G1,s);
	auto d2=Dijkstra(G2,t);

	vector<lint> ans(n);
	lint mn=INF;
	for(int u=n-1;u>=0;u--){
		mn=min(mn,d1[u]+d2[u]);
		ans[u]=1000000000000000-mn;
	}
	rep(u,n) printf("%lld\n",ans[u]);

	return 0;
}
