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 <vector>
#include <queue>
#include <utility>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> P;
ll n,m,s,t,u,v,a,b;
ll Cy[100010],Cs[100010];
ll have = 1e15,inf = 1e18;
vector<vector<P>> Y(100010),S(100010);

int main(){
	cin >> n >> m >> s >> t;
	for(int i=0;i<m;i++){
		cin >> u >> v >> a >> b;
		Y[u].push_back(P(a,v));
		Y[v].push_back(P(a,u));
		S[u].push_back(P(b,v));
		S[v].push_back(P(b,u));
	}
	priority_queue<P> Q;
	Q.push(P(0,s));
	for(int i=0;i<=n;i++){
		Cy[i] = inf;
	}
	Cy[s] = 0;
	while(!Q.empty()){
		P t = Q.top();
		ll w = t.second;
		Q.pop();
		if(Cy[w] < t.first) continue;
		for(int i=0;i<Y[w].size();i++){
			P e = Y[w][i];
			if(Cy[e.second] > Cy[w] + e.first){
				Cy[e.second] = Cy[w] + e.first;
				Q.push(P(-Cy[e.second],e.second));
			}
		}
	}
	Q.push(P(0,t));
	for(int i=0;i<=n;i++){
		Cs[i] = inf;
	}
	Cs[t] = 0;
	while(!Q.empty()){
		P t = Q.top();
		ll w = t.second;
		Q.pop();
		if(Cs[w] < t.first) continue;
		for(int i=0;i<S[w].size();i++){
			P e = S[w][i];
			if(Cs[e.second] > Cs[w] + e.first){
				Cs[e.second] = Cs[w] + e.first;
				Q.push(P(-Cs[e.second],e.second));
			}
		}
	}
	ll A[100010] = {0};
	for(int i=1;i<=n;i++){
		A[i] = Cy[i] + Cs[i];
	}
	for(int i=n-1;i>=1;i--){
		A[i] = min(A[i],A[i+1]);
	}
	//sort(T.begin(),T.end());
	//int id = 1;
	for(int i=1;i<=n;i++) cout << have-A[i] << endl;
}