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;
typedef pair<ll,ll> P;

ll N,M,S,T;
vector<vector<P>> X;
vector<vector<P>> Y;
ll INF = 1e18;

int main(){
    cin >> N >> M >> S >> T;
    S--, T--;
    X.resize(N);
    Y.resize(N);
    for(int i=0;i<M;i++){
        ll u,v,a,b;
        cin >> u >> v >> a >> b;
        X[u-1].push_back({v-1,a});
        X[v-1].push_back({u-1,a});
        Y[u-1].push_back({v-1,b});
        Y[v-1].push_back({u-1,b});
    }
    vector<ll> x(N,INF);
    vector<ll> y(N,INF);
    priority_queue<P,vector<P>,greater<P>> pq;
    pq.push({0,S});
    while(!pq.empty()){
        auto t = pq.top();
        ll c = t.first;
        ll n = t.second;
        pq.pop();
        if(x[n] != INF) continue;
        x[n] = c;
        for(auto &e : X[n]){
            if(x[e.first] != INF) continue;
            pq.push({e.second + c, e.first});
        }
    }
    pq.push({0,T});
    while(!pq.empty()){
        auto t = pq.top();
        ll c = t.first;
        ll n = t.second;
        pq.pop();
        if(y[n] != INF) continue;
        y[n] = c;
        for(auto &e : Y[n]){
            if(y[e.first] != INF) continue;
            pq.push({e.second + c, e.first});
        }
    }
    vector<ll> ans;
    ans.push_back(-1);
    for(int i=0;i<N;i++){
        ll tmp = 1e15 - x[N-i-1] - y[N-i-1];
        ans.push_back(max(ans.back(),tmp));
    }
    for(int i=0;i<N;i++){
        cout << ans[N-i] << endl;
    }
    return 0;
}