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
// 円とスヌークについてdijkstraを2本張る
#include<iostream>
#include<vector>
#include<queue>
using namespace std;

typedef long long ll;
const ll INF = 1ll<<60;

struct Edge{int to; ll cost;};

int n, m, s, t, u, v, a, b;

void dijkstra(int from, vector<vector<Edge>> &path, ll *dist){
    dist[from] = 0;
    priority_queue<pair<ll, int>> pq;

    pq.push({dist[from], from});

    while(!pq.empty()){
        pair<ll, int> now = pq.top();   pq.pop();
        ll c = -now.first;
        int pos = now.second;
        if(c > dist[pos])    continue;
        for(Edge e : path[pos]){
            if(dist[e.to] > c + e.cost){
                dist[e.to] = c + e.cost;
                pq.push({-dist[e.to], e.to});
            }
        }
    }

    return;
}

int main(){
    cin >> n >> m >> s >> t;
    s--, t--;
    vector<vector<Edge>> yen(n), snu(n);

    for(int i = 0; i < m; i++){
        cin >> u >> v >> a >> b;
        u--, v--;
        yen[u].push_back(Edge{v,a});
        yen[v].push_back(Edge{u,a});
        snu[u].push_back(Edge{v,b});
        snu[v].push_back(Edge{u,b});
    }

    ll dijky[n], dijks[n];
    for(int i = 0; i < n; i++)  dijky[i] = dijks[i] = INF;

    dijkstra(s, yen, dijky);
    dijkstra(t, snu, dijks);

    ll ans[n];
    ans[n-1] = dijky[n-1]+dijks[n-1];

    for(int i = n-2; i >= 0; i--){
        ans[i] = min(ans[i+1], dijky[i]+dijks[i]);
    }

    for(int i = 0; i < n; i++){
        cout << 1000000000000000-ans[i] << endl;
    }
    return 0;
}