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;
typedef long long ll;
#define rep(i, srt, end) for (long long i = (srt); i < (long long)(end); i++)

/*
Arguments:
- G   : 隣接リスト {destination edge, cost}
- d   : srtからの距離 <- これを変更する
- srt : 始点
*/

void dijkstra(vector<vector<pair<ll,ll>>> G, vector<ll> &d, ll srt){
    priority_queue<pair<ll,ll>, vector<pair<ll,ll>>, greater<pair<ll,ll>>> que;
    que.push({0, srt});
    d[srt] = 0;
    while(!que.empty()){
        auto q = que.top();
        que.pop();
        if(q.first > d[q.second]) continue;
        for(auto g : G[q.second]){
            if(d[g.first] <= q.first + g.second) continue;
            d[g.first] = q.first + g.second;
            que.push({d[g.first], g.first});
        }
    }
}

int main(){
    ll n, m, s, t;
    cin >> n >> m >> s >> t;
    s--; t--;
    vector<vector<pair<ll,ll>>> Gs(n), Gt(n);
    rep(i, 0, m) {
        ll u, v, a, b;
        cin >> u >> v >> a >> b;
        u--; v--;
        Gs[u].push_back({v, a});
        Gs[v].push_back({u, a});
        Gt[u].push_back({v, b});
        Gt[v].push_back({u, b});
    }
    vector<ll> ds(n, 1e18), dt(n, 1e18);
    dijkstra(Gs, ds, s);
    dijkstra(Gt, dt, t);
    vector<pair<ll,ll>> cands(n);
    rep(i, 0, n) cands[i] = {ds[i] + dt[i], i+1};
    sort(cands.begin(), cands.end());
    ll itr = 0;
    rep(i, 0, n){
        while(cands[itr].second <= i) itr++;
        cout << 1000000000000000 - cands[itr].first << endl;
    }
    return 0;
}