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>

using namespace std;

const long long INF = (1LL << 60);

vector<long long> dijkstra(const vector<vector<pair<int,int>>>& g, int s){
	const int n = g.size();
	vector<long long> res(n, INF);
	priority_queue<pair<int,int>, vector<pair<int,int>>, greater<pair<int,int>>> qu;
	qu.emplace(0, s);
	res[s] = 0;
	while(!qu.empty()){
		auto p = qu.top(); qu.pop();
		for(const auto& next : g[p.second]){
			int nextPos = next.second;
			long long nextDist = res[p.second] + next.first;
			if(res[nextPos] > nextDist){
				res[nextPos] = nextDist;
				qu.emplace(nextDist, nextPos);
			}
		}
	}
	return res;
}

int main(){
	int n, m, s, t;
	while(cin >> n >> m >> s >> t){
		--s; --t;
		vector<vector<pair<int,int>>> g1(n), g2(n);
		for(int i=0;i<m;i++){
			int u, v, a, b; cin >> u >> v >> a >> b;
			--u; --v;
			g1[u].emplace_back(a, v);
			g1[v].emplace_back(a, u);
			g2[u].emplace_back(b, v);
			g2[v].emplace_back(b, u);
		}
		auto dist1 = dijkstra(g1, s);
		auto dist2 = dijkstra(g2, t);
		vector<long long> res(n, INF);
		res[n-1] = dist1[n-1] + dist2[n-1];
		for(int i=n-2;i>=0;i--){
			res[i] = min(res[i+1], dist1[i]+dist2[i]);
		}
		for(auto& t : res) cout << 1000000000000000 - t << endl;
	}
}
