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;
using ll=long long;
using ld=long double;
using P=pair<ll,ll>;
#define MOD 1000000007ll
#define INF 1000000000ll
#define EPS 1e-10
#define FOR(i,n,m) for(ll i=n;i<(ll)m;i++)
#define REP(i,n) FOR(i,0,n)
#define DUMP(a) REP(d,a.size()){cout<<a[d];if(d!=a.size()-1)cout<<" ";else cout<<endl;}
#define ALL(v) v.begin(),v.end()
#define UNIQUE(v) sort(ALL(v));v.erase(unique(ALL(v)),v.end());
#define pb push_back

ll n,m,s,t;
vector<ll> dijkstra(vector<vector<P>>& g, ll sp) {
	vector<ll> dir(n,INF*INF);
	dir[sp]=0;
	priority_queue<P,vector<P>,greater<P>> q;
	q.push(P(0,sp));
	while(!q.empty()) {
		ll d=q.top().first;
		ll p=q.top().second;
		q.pop();
		if(dir[p]!=d) continue;
		REP(i,g[p].size()) {
			ll np=g[p][i].first;
			ll cost=g[p][i].second;
			if(dir[np]>d+cost) {
				dir[np]=d+cost;
				q.push(P(dir[np],np));
			}
		}
	}
	return dir;
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin>>n>>m>>s>>t;
	s--; t--;
	vector<vector<P>> gy(n);
	vector<vector<P>> gs(n);
	REP(i,m) {
		ll u,v,a,b;
		cin>>u>>v>>a>>b;
		u--;
		v--;
		gy[u].pb(P(v,a));
		gy[v].pb(P(u,a));
		gs[u].pb(P(v,b));
		gs[v].pb(P(u,b));
	}
	vector<ll> ycost=dijkstra(gy,s);
	vector<ll> scost=dijkstra(gs,t);
	const ll init=1000000000000000LL;
	priority_queue<P,vector<P>,greater<P>> q;
	REP(i,n) {
		q.push(P(ycost[i]+scost[i],i));
	}
	REP(i,n) {
		while(q.top().second<i) {
			q.pop();
		}
		cout<<init-q.top().first<<endl;
	}
}
