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;
const ll INF=1e15;

template<typename T>
struct Dijkstra{
    const T inf=numeric_limits<T>::max();
    using P=pair<T,int>;
    int n;
    vector<vector<pair<int,T>>> G;
    vector<T> d;
    Dijkstra(int n):n(n),G(n),d(n){}
    void add_edge(int u,int v,T w){
        G[u].emplace_back(v,w);
    }
    vector<T> build(int s){
        fill(d.begin(),d.end(),inf);
        d[s]=0;
        priority_queue<P,vector<P>,greater<P>> pq;
        pq.emplace(d[s],s);
        while(!pq.empty()){
            P p=pq.top(); pq.pop();
            int v=p.second;
            if (d[v]<p.first) continue;
            for (auto &e:G[v]){
                int u=e.first; T c=e.second;
                if (d[v]+c<d[u]){
                    d[u]=d[v]+c;
                    pq.emplace(d[u],u);
                }
            }
        }
        return d;
    }
};

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    int n,m,s,t; cin >> n >> m >> s >> t;
    Dijkstra<ll> Ds(n),Dt(n);
    for (int i=0;i<m;++i){
        int u,v; ll a,b; cin >> u >> v >> a >> b; --u,--v;
        Ds.add_edge(u,v,a); Ds.add_edge(v,u,a);
        Dt.add_edge(u,v,b); Dt.add_edge(v,u,b);
    }
    vector<ll> S=Ds.build(--s),T=Dt.build(--t);
    vector<ll> ans(n+1,INF);
    for (int i=n-1;i>=0;--i){
        ans[i]=min(ans[i+1],S[i]+T[i]);
    }
    for (int i=0;i<n;++i) cout << INF-ans[i] << '\n';
}