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 <algorithm>
#include <functional>
#include <queue>
#define int long long
#define rep(i, n) for(i = 0; i < n; i++)
using namespace std;

int INF = 1e+15;
int n, m, s, t;
vector<int> et[100000];
vector<int> ec[2][100000];	//円, すぬーく
int minCost[2][2][100000];	//始点{S, T}, {円, すぬーく}, 頂点
int sumCost[100000];

void bfs(int s, int type, int minCost[]) {
	typedef pair<int, int> P;
	static priority_queue<P, vector<P>, greater<P> > que;
	int i;
	
	rep(i, n) minCost[i] = INF;
	
	que.push(P(0, s));
	minCost[s] = 0;
	
	while (!que.empty()) {
		P now = que.top(); que.pop();
		int cst = now.first;
		int v = now.second;
		
		rep(i, et[v].size()) {
			int nv = et[v][i];
			int ncst = cst + ec[type][v][i];
			if (minCost[nv] > ncst) {
				minCost[nv] = ncst;
				que.push(P(ncst, nv));
			}
		}
	}
}

signed main() {
	int i;
	
	cin >> n >> m >> s >> t;
	s--; t--;
	rep(i, m) {
		int u, v, a, b;
		cin >> u >> v >> a >> b;
		u--; v--;
		et[u].push_back(v);
		et[v].push_back(u);
		ec[0][u].push_back(a);
		ec[0][v].push_back(a);
		ec[1][u].push_back(b);
		ec[1][v].push_back(b);
	}
	
	bfs(s, 0, minCost[0][0]);
	bfs(t, 1, minCost[1][1]);
	
	rep(i, n) {
		//s -> i -> t
		sumCost[i] = minCost[0][0][i] + minCost[1][1][i];
	}
	
	int mmin = INF;
	vector<int> ans;
	for (i = n - 1; i >= 0; i--) {
		mmin = min(mmin, sumCost[i]);
		ans.push_back(mmin);
	}
	
	for (i = n - 1; i >= 0; i--) {
		cout << INF - ans[i] << endl;
	}
	return 0;
}