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<cstdio>
#include<iostream>
#include<cstring>
#include<vector>
#include<queue>
using namespace std;

#define ran 101101
#define i64 long long

int n, m, s, t;
vector<pair<int, int> > e[ran], _e[ran];
i64 d[2][ran], ans[ran];
priority_queue<pair<i64, int> > q;

void dij(int st, int id) {
	while(!q.empty()) q.pop();
	q.push(make_pair(0LL, st));
	int todo = n;
	memset(d[id], -1, sizeof d[id]);
	while(todo > 0 && !q.empty()) {
		pair<i64, int> p = q.top(); q.pop();
		i64 dist = -p.first;
		int cur = p.second;
		if (d[id][cur]!= -1LL) continue;
		
		todo--;
		d[id][cur] = dist;
		for(int i=0;i<(int)e[cur].size();i++) {
			int nxt = e[cur][i].first;
			if (d[id][nxt] == -1LL)
				q.push(make_pair(-(dist + e[cur][i].second), nxt));
		}
	}
}

int main() {
	cin >> n >> m >> s >> t;
	for(int i=0;i<m;i++) {
		int x, y, va, vb;
		cin >> x >> y >> va >> vb;
		e[x].push_back(make_pair(y, va));
		e[y].push_back(make_pair(x, va));
		_e[x].push_back(make_pair(y, vb));
		_e[y].push_back(make_pair(x, vb));
	}
	dij(s, 0);
	for(int i=1;i<=n;i++) e[i] = _e[i];
	dij(t, 1);
	long long mi = d[0][n] + d[1][n];
	for(int i=n;i>0;i--) {
		mi = min(mi, d[0][i] + d[1][i]);
		ans[i] = mi;
	}
	for(int i=1;i<=n;i++) cout<<1000000000000000LL - ans[i]<<endl;
	
	return 0;
}