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 ll = long long;
using P = pair<ll, int>;
const ll MOD = 1000000007;

struct edge{
    int to;
    ll cost;
    edge(int to, ll cost): to(to), cost(cost){}
};

vector<ll> dijkstra(vector<vector<edge>>& E, int start=0){
    vector<ll> d(E.size(), 1LL << 60);
    priority_queue<P, vector<P>, greater<P>> que;
    d[start] = 0;
    que.push(P(0, start));
    while(que.size()){
        P p = que.top(); que.pop();
        if(d[p.second] < p.first) continue;
        for(edge e: E[p.second]){
            ll nc = p.first + e.cost;
            if(d[e.to] > nc){
                d[e.to] = nc;
                que.push(P(nc, e.to));
            }
        }
    }
    return d;
}

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

    vector<vector<edge>> G_a(n), G_b(n);
    for(int i=0;i<m;i++){
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        u--;
        v--;
        G_a[u].emplace_back(v, a);
        G_a[v].emplace_back(u, a);
        G_b[u].emplace_back(v, b);
        G_b[v].emplace_back(u, b);
    }

    auto dist_a = dijkstra(G_a, s);
    auto dist_b = dijkstra(G_b, t);
    vector<ll> dist_ab(n);
    for(int i=0;i<n;i++){
        dist_ab[i] = dist_a[i] + dist_b[i];
    }
    for(int i=n-2;i>=0;i--){
        dist_ab[i] = min(dist_ab[i], dist_ab[i + 1]);
    }
    for(int i=0;i<n;i++){
        cout << 1000000000000000LL - dist_ab[i] << endl;
    }

    return 0;
}