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>
#define INF (long long) (1e15)
using namespace std;

vector<long long> dijkstra(vector<vector<pair<int, long long>>>& next, int s) {
    int n = next.size();
    vector<long long> dist(n, INF);
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> pq;
    dist[s] = 0;
    pq.push(make_pair(0, s));
    while (pq.size()) {
        int u = pq.top().second;
        pq.pop();
        for (auto t: next[u]) {
            int v = t.first;
            long long alt = dist[u] + t.second;
            if (dist[v] > alt) {
                dist[v] = alt; 
                pq.push(make_pair(alt, v));
            }
        }
    }
    return dist;
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    int n, m, s, t;
    cin >> n >> m >> s >> t;
    s--; t--;
    vector<vector<pair<int, long long>>> A(n), B(n);
    for (int i = 0; i < m; i++) {
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        u--; v--;
        A[u].push_back(make_pair(v, a));
        A[v].push_back(make_pair(u, a));
        B[u].push_back(make_pair(v, b));
        B[v].push_back(make_pair(u, b));
    }
    vector<long long> fromS = dijkstra(A, s), fromT = dijkstra(B, t), ans(n + 1, INF);
    for (int i = n - 1; i >= 0; i--) {
        ans[i] = min(ans[i + 1], fromS[i] +  fromT[i]);
    }
    for (int i = 0; i < n; i++) {
        cout << INF - ans[i] << endl;
    }

    

}
