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;
#define int long long
int inf=1e18;

struct graph{

struct edge{int to,cst;};
struct pos{int i,d;
	pos(){};
	pos(int i_,int d_){i=i_,d=d_;};
	bool operator<(const pos& r)const{return d>r.d;}
};


vector<vector<edge>>g;
vector<bool>used;
int n;

vector<int>dist;
priority_queue<pos> pq;

	graph(){};
	graph(int n_){
	n=n_;
	g.resize(n);
	used.assign(n,false);
	dist.resize(n,inf);
	};

	void add_edge(int x,int y,int cst){
		g[x].push_back(edge({y,cst}));
		g[y].push_back(edge({x,cst}));
	}

	void set_dist(int y){
		dist[y]=0;
		bool first=true;
		 pq.push(pos(y,0));
		while(!pq.empty()){
			pos x=pq.top();pq.pop();
			if(dist[x.i]<x.d)continue;
			first=false;
			for(auto&& w:g[x.i]){
				if(dist[w.to]> x.d+w.cst){
					dist[w.to]=x.d+w.cst;
					pq.push(pos(w.to,dist[w.to]));
				}
			} 
		}	
	}

};


signed main(){
	int n,m,s,t;cin>>n>>m>>s>>t;s--;t--;
	graph g1(n),g2(n);
	for(int i=0;i<m;i++){
		int x,y,a,b;cin>>x>>y>>a>>b;x--;y--;
		g1.add_edge(x,y,a);
		g2.add_edge(x,y,b);
	}
	g1.set_dist(s);
	g2.set_dist(t);	
	int money=1e15;
	int cst=inf;
	vector<int>cs(n);
	for(int i=n-1;i>=0;i--){
		cst=min(cst,g1.dist[i]+g2.dist[i]);
		cs[i]=cst;
	}
	for(auto&& w:cs)cout<<money-w<<endl;
}
