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;

#define rep(i, a, b) for(int i = a; i < (b); ++i)
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int N = 1e5;
const ll WALLET = 1e15;
const ll INF = 1e18;

int n, m, s, t;
vector<pii> conn[2][N];
ll cost[2][N], tot[N];

void dijkstra(int start, int cur) {
	rep(i, 0, n) cost[cur][i] = INF;
	cost[cur][start] = 0;
	
	set<pair<ll, ll>> q;
	q.insert({0, start});
	
	while (!q.empty()) {
		int v = q.begin()->second;
		q.erase(q.begin());
		for (auto u : conn[cur][v]) {
			ll to = u.first, price = u.second;
			if (cost[cur][v] + price < cost[cur][to]) {
				q.erase({cost[cur][to], to});
				cost[cur][to] = cost[cur][v] + price;
				q.insert({cost[cur][to], to});
			}
		}
	}
}

int main() {
	cin.tie(0)->sync_with_stdio(0);
	cin >> n >> m >> s >> t;
	--s, --t;
	rep(i, 0, m) {
		ll u, v, a, b;
		cin >> u >> v >> a >> b;
		--u, -- v;
		conn[0][u].push_back({v, a});
		conn[1][u].push_back({v, b});
		conn[0][v].push_back({u, a});
		conn[1][v].push_back({u, b});
	}
	
	dijkstra(s, 0);
	dijkstra(t, 1);
	
	set<pair<ll, ll>> best;
	rep(i, 0, n) {
		tot[i] = cost[0][i] + cost[1][i];
		best.insert({tot[i], i});
	}
	
	rep(i, 0, n) {
		cout << WALLET - best.begin()->first << '\n';
		best.erase({tot[i], i});
	}
}
