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 <map>
#include <set>
#include <queue>
#include <string>
#include <iomanip>
#include <algorithm>
#include <cmath>
#include <stdio.h>
using namespace std;
#define int long long
int MOD = 1000000007;
int N, M;
int INF = (int)1 << 60;
vector<int> solve(vector<vector<int> > &g , vector<vector<int> > &A, int s) {
	vector<int> res(N, INF);
	priority_queue<pair<int,int > > pq;
	res[s] = 0;
	pq.emplace(0, s);
	while (pq.size()) {
		int a = pq.top().second; pq.pop();
		for (int i = 0; i < g[a].size(); i++) {
			if (res[g[a][i]] > res[a] + A[a][i]) {
				res[g[a][i]] = res[a] + A[a][i];
				pq.emplace(-res[g[a][i]], g[a][i]);
			}
		}
	}
	return res;
}
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	int S, T;
	cin >> N >> M >> S >> T;
	S--;
	T--;
	vector<vector<int>  >g(N);
	vector<vector<int>  >a(N);
	vector<vector<int>  >b(N);
	int u, v, aa, bb;
	for (int i = 0; i < M; i++) {
		cin >> u >> v >> aa >> bb;
		u--;
		v--;
		g[u].push_back(v);
		g[v].push_back(u);
		a[u].push_back(aa);
		a[v].push_back(aa);
		b[u].push_back(bb);
		b[v].push_back(bb);
	}
	vector<int> A, B;
	A = solve(g, a, S);
	B = solve(g, b, T);
	vector<int> res(N);
	
	res[N - 1] = A[N - 1] + B[N - 1]; 
	for (int i = N - 2; i >= 0; i--) {
		res[i] = min(res[i + 1], A[i] + B[i]);
	}
	int z = (int)(1e15);
	for (int i = 0; i < N; i++) {
		cout << z - res[i] << endl;
	}
}