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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
const ll inf = 1e18;
vector<pair<int, int> > e1[N];
vector<pair<int, int> > e2[N];
ll a[N], b[N];
bool flag[N];
int n, m, s, t;
void spfa(int s, int t, vector<pair<int, int> > e[], ll d[]) {
	memset(flag, 0, sizeof(flag));
	for (int i = 1; i <= n; i++) d[i]=inf;
	queue<int> q;
	q.push(s);
	d[s] = 0;
	flag[s] = true;
	while (!q.empty()) {
		int u = q.front();
		q.pop();
		flag[u] = false;
		for (auto t : e[u]) {
			int v = t.first;
			int w = t.second;
			if (d[u] + w < d[v]) {
				d[v] = d[u] + w;
				if (!flag[v]) {
					flag[v] = true;
					q.push(v);
				}
			}
		}
	}
}

int main() {
	scanf("%d%d%d%d", &n, &m, &s, &t);
	for (int i = 0; i < m; i++) {
		int u, v, a, b;
		scanf("%d%d%d%d", &u, &v, &a, &b);
		e1[u].emplace_back(v, a);
		e1[v].emplace_back(u, a);
		e2[u].emplace_back(v, b);
		e2[v].emplace_back(u, b);
	}
	spfa(s, t, e1, a);
	spfa(t, s, e2, b);
	set<pair<ll, int> > S;
	for (int i = 1; i <= n; i++) S.insert(make_pair(a[i]+b[i], i));
	ll sum = 1e15;
	for (int i = 0; i < n; i++) {
		while ((*S.begin()).second <= i) S.erase(S.begin());
		printf("%lld\n", sum-(*S.begin()).first);
	}
	return 0;
}
