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;
using Int = long long;
using A2 = array<int, 2>;
#define printa(A) for (auto a : A) cout << a << ' '; cout << endl;
template<class T> using priority_queue_min = priority_queue<T, vector<T>, greater<T>>;
void dijkstra(vector<vector<pair<int, A2>>>& G, int s, int idx, vector<Int>& out)
{
    priority_queue_min<pair<Int, int>> Q;
    Q.emplace(0, s);
    vector<bool> visited(G.size());
    while (!Q.empty()) {
        Int d; int i; tie(d, i) = Q.top(); Q.pop();
        if (visited[i]) continue;
        visited[i] = true;
        out[i] = d;
        for (auto &p : G[i]) {
            if (!visited[p.first]) {
                Q.emplace(d + p.second[idx], p.first);
            }
        }
    }
}
int main()
{
    int N, M, S, T; cin >> N >> M >> S >> T; S--, T--;
    vector<vector<pair<int, A2>>> G(N);
    while (M--) {
        int u, v, a, b; cin >> u >> v >> a >> b; u--, v--;
        G[u].emplace_back(v, A2{a, b});
        G[v].emplace_back(u, A2{a, b});
    }
    vector<Int> B(N), C(N);
    dijkstra(G, S, 0, B);
    dijkstra(G, T, 1, C);
    vector<Int> cost(N);
    cost[N-1] = B[N-1] + C[N-1];
    for (int i = N-2; i >= 0; i--) {
        cost[i] = min(cost[i+1], B[i] + C[i]);
    }
    for (int i = 0; i < N; i++) {
        cout << (Int)1e15 - cost[i] << '\n';
    }
    return 0;
}