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;
using pi = pair<ll,ll>;
const ll MOD = 1e9 + 7;
const ll INF = 1e18;

void dijkstra(vector<vector<pair<ll,ll>>> &G, ll n, vector<ll> &dist, ll start) {
    priority_queue<pi, vector<pi>, greater<pi>> pq;
    dist[start] = 0;
    vector<ll> used(n+1,0);
    pq.push({0, start});
    while(!pq.empty()) {
        ll d = pq.top().first, v = pq.top().second;
        pq.pop();
        if(used[v])continue;
        used[v] = 1;
        if(dist[v]<d){
            continue;
        }
        for(pi e : G[v]) {
            ll to = e.first, cost = e.second;
            if(dist[v] + e.second < dist[to]) {
                dist[to] = dist[v] + cost;
                pq.push({dist[to], to});
            }
        }
    }
}

int main() {
    ll n, m, i, j, k,s,t;
    cin >> n >> m >> s >> t;
    vector<vector<pair<ll,ll>>>G1(n+1),G2(n+1);
    for(i=0;i<m;i++){
        ll u,v,a,b;
        cin >> u >> v >> a >> b;
        G1[u].push_back({v,a});
        G1[v].push_back({u,a});
        G2[u].push_back({v,b});
        G2[v].push_back({u,b});
    }
    vector<ll> dista(n + 2, INF), distb(n + 2, INF);
    dijkstra(G1,n,dista,s);
    dijkstra(G2,n,distb,t);
    ll ret = INF;
    vector<ll> v;
    ll money = 1000000000000000;
    for(i=n-1;i>=0;i--){
        ret = min(dista[i+1]+distb[i+1],ret);
        v.push_back(ret);
    }
    reverse(v.begin(),v.end());
    for(ll x:v){
        cout << money-x << endl;
    }
    return 0;
}