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 <iostream>
#include <vector>
#include <queue>
#include <utility>
#include <algorithm>
#define ll long long
using namespace std;

ll MX=1e15;

void dijkstra(int start, vector<vector<int>> &g, vector<vector<ll>> &cost, vector<ll> &d){
    priority_queue<pair<ll, int>, vector<pair<ll, int>>, greater<pair<ll, int>>> pq;
    int now=start, cnt=0;
    while(cnt<d.size()-2){
        for(int i=0; i<g[now].size(); ++i){
            if(d[g[now][i]]==MX){
                pq.push(make_pair(d[now]+cost[now][i], g[now][i]));
            }
        }
        while(d[pq.top().second]<MX) pq.pop();
        now=pq.top().second;
        d[now]=pq.top().first;
        pq.pop();
        ++cnt;
    }
}

int main() {
	int n, m, s, t;
	cin >> n >> m >> s >> t;
	vector<vector<int>> graph(n+1);
	vector<vector<ll>> yen(n+1), snuuk(n+1);
	for(int i=0; i<m; ++i){
	    int u, v;
	    ll a, b;
	    cin >> u >> v >> a >> b;
	    graph[u].push_back(v);
	    yen[u].push_back(a);
	    snuuk[u].push_back(b);
	    graph[v].push_back(u);
	    yen[v].push_back(a);
	    snuuk[v].push_back(b);
	}
	vector<ll> d_yen(n+1, MX), d_snuuk(n+1, MX);
	d_yen[s]=0;
	d_snuuk[t]=0;
	dijkstra(s, graph, yen, d_yen);
	dijkstra(t, graph, snuuk, d_snuuk);
	vector<ll> ans(n+1, 0);
	for(int i=n; i>0; --i) ans[i-1]=max(ans[i], MX-(d_yen[i]+d_snuuk[i]));
	for(int i=0; i<n; ++i) cout << ans[i] << endl;
	return 0;
}
