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;
#define int long long
using P = pair<int,int>;

struct edge{int to,idx;};
vector<edge> G[100100];

vector<int> dijkstra(int s,vector<int>& cost){
    vector<int> dist(100100,1e18); dist[s] = 0;
    priority_queue<P,vector<P>,greater<P>> Q;
    Q.push({0,s});
    while(!Q.empty()){
        P p = Q.top(); Q.pop();
        if(dist[p.second] < p.first) continue;
        for(auto &e : G[p.second]){
            if(dist[e.to] > dist[p.second] + cost[e.idx]){
                dist[e.to] = dist[p.second] + cost[e.idx];
                Q.push({dist[e.to],e.to});
            }
        }
    }
    return dist;
}

signed main(){
    int N,M,S,T;
    cin>>N>>M>>S>>T;
    vector<int> A(M),B(M);
    for(int i=0;i<M;++i){
        int u,v; cin>>u>>v>>A[i]>>B[i];
        G[u].push_back({v,i});
        G[v].push_back({u,i});
    }

    auto yen = dijkstra(S,A);
    auto snuuk = dijkstra(T,B);

    int ans = 0;
    vector<int> res;

    for(int i=N;i>0;--i){
        ans = max(ans,(int)1e15-yen[i]-snuuk[i]);
        res.push_back(ans);
    }

    for(int i=N-1;i>=0;--i) cout<<res[i]<<endl;
}