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;

struct edge {
    int to;
    int cost;
};

const long long INF = 1e18;

typedef pair<long long, long long> P;
vector<long long> dijk(vector<vector<edge>> &g, int n, int s) {
    vector<long long> d(n);
    priority_queue<P, vector<P>, greater<P>> q;
    fill(d.begin(), d.end(), INF);
    d[s] = 0;

    q.push({0, s});

    while(!q.empty()) {
        auto p = q.top(); q.pop();
        auto v = p.second;
        if (d[v] < p.first) continue;
        for (auto &e : g[v]) {
            if (d[e.to] > d[v] + e.cost) {
                d[e.to] = d[v] + e.cost;
                q.push({d[e.to], e.to});
            }
        }
    }
    return d;
}

int main() {
    int n, m, s, t;
    cin >> n >> m >> s >> t;

    vector<vector<edge>> G1(n), G2(n);
    for (int i = 0; i < m; ++i) {
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        --u, --v;
        G1[u].push_back({v, a});
        G1[v].push_back({u, a});
        G2[u].push_back({v, b});
        G2[v].push_back({u, b});
    }
    auto d1 = dijk(G1, n, s - 1);
    auto d2 = dijk(G2, n, t - 1);

    priority_queue<P> q;
    const long long X = 1e15;
    for (int i = 0; i < n; ++i) {
        q.push({X - (d1[i] + d2[i]), i});
    }
    for (int i = 0; i < n; ++i) {
        while (q.top().second < i) q.pop();
        cout << q.top().first << endl;
    }

    return 0;
}