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 rep(i,n) for(int i=0; i<(n); i++)
const int MX = 100005;
const long long INF = 1e15;
using P = pair<long long, int>;
int n, m, s[2], u, v, a[2], from, to;
long long dist[2][MX], ac, cost, ans[MX];
vector<P> G[2][MX];
priority_queue<P, vector<P>, greater<P>> qu;
void dijkstra(int s, int f){
    fill(dist[f], dist[f]+n, INF);
    qu.push(P(dist[f][s] = 0,s));
    while(!qu.empty()){
        tie(ac,from) = qu.top(); qu.pop();
        if(dist[f][from] == ac) for(P it: G[f][from]){
            tie(cost, to) = it; cost += ac;
            if(dist[f][to] > cost) qu.push(P(dist[f][to] = cost, to));
        }
    }
}
int main(){
    scanf("%d%d%d%d", &n, &m, s, s+1);
    s[0]--; s[1]--;
    rep(i,m){
        scanf("%d%d%d%d", &u, &v, a, a+1);
        u--; v--;
        rep(j,2){
            G[j][u].push_back(P(a[j],v));
            G[j][v].push_back(P(a[j],u));
        }
    }
    rep(i,2) dijkstra(s[i],i);
    for(int i = n-1; i >= 0; i--) ans[i] = max(ans[i+1], INF - dist[0][i] - dist[1][i]);
    rep(i,n) printf("%lld\n", ans[i]);
}
