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 <string>
#include <list>
#include <queue>
#include <algorithm>
#define rep(i, n) for(i = 0; i < (n); i++)
#define chmax(x, y) x = max(x, y)
#define chmin(x, y) x = min(x, y)
#define MOD 1000000007
#define PI 3.14159265358979323846
#define INF 1ll << 60

using namespace std;
typedef long long ll;
typedef pair<ll, int> pp;

typedef struct {
	int to;
	ll cost;
} edge;

void Dijk(int s, vector<vector<edge>> &d, vector<ll> &ans) {
	int i;
	priority_queue<pp, vector<pp>, greater<pp>> que;
	que.push(pp(0, s));
	while (!que.empty()) {
		int n = que.top().second;
		ll c = que.top().first;
		que.pop();
		if (ans[n] == INF) {
			ans[n] = c;
			vector<edge>& e = d[n];
			rep(i, e.size()) {
				if (ans[e[i].to] == INF) {
					que.push(pp(e[i].cost + c, e[i].to));
				}
			}
		}
	}
}

int main(void) {
	int num, i, m, s, g, x, y, c, d;
	ll sum = 1e15;
	cin >> num >> m >> s >> g;
	vector<vector<edge>> a(num + 1), b(num + 1);
	vector<ll> fir(num + 1, INF), lat(num + 1, INF), ans(num + 1);
	priority_queue<ll> que;
	rep(i, m) {
		scanf("%d%d%d%d", &x, &y, &c, &d);
		a[x].push_back({ y, c });
		a[y].push_back({ x, c });
		b[x].push_back({ y, d });
		b[y].push_back({ x, d });
	}
	Dijk(s, a, fir);
	Dijk(g, b, lat);
	for (i = num; i > 0; i--) {
		que.push(sum - fir[i] - lat[i]);
		ans[i] = que.top();
	}
	for (i = 1; i <= num; i++)
		cout << ans[i] << "\n";
	return 0;
}