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;

void dijk(int s, vector<vector<pair<int,int>>> &E, vector<long long> &D)
{
    int N=D.size(),vis[101010];
    for (int i = 0; i < N; ++i)
    {
        D[i]=10000000000000000LL;
    }
    for (int i = 0; i < N; ++i)
    {
        vis[i]=0;
    }
 
    priority_queue<pair<long long, int>, vector<pair<long long, int>>, greater<pair<long long, int>>> que;
 
    D[s] = 0;
    que.push({ 0, s });
 
    while (!que.empty()) 
    {
        auto q = que.top(); que.pop();
 
        long long cst = q.first;
        int cu = q.second;
 
        if (vis[cu]) continue;
        vis[cu] = 1;
 
        for(auto &p:E[cu]) 
        {
            long long cst2 = cst + p.second;
            int to = p.first;
 
            if (D[to]>cst2)
            {
                D[to]=cst2;
                que.push({D[to],to});
            }
        }
    }
}

int main(int argc, char const *argv[])
{
	int n,m,s,t;
	cin>>n>>m>>s>>t;--s;--t;
	vector<vector<pair<int,int>>> edge1(n),edge2(n);
	int u,v,a,b;
	for (int i = 0; i < m; ++i)
	{
		cin>>u>>v>>a>>b;--u;--v;
		edge1[u].push_back({v,a});
		edge1[v].push_back({u,a});
		edge2[u].push_back({v,b});
		edge2[v].push_back({u,b});
	}
	vector<long long> d1(n),d2(n);
	dijk(s,edge1,d1);
	dijk(t,edge2,d2);
	long long ans[n];
	ans[n-1]=d1[n-1]+d2[n-1];
	for (int i = n-2; i >= 0; --i)
	{
		ans[i]=min(ans[i+1],d1[i]+d2[i]);
	}
	for (int i = 0; i < n; ++i)
	{
		cout<<1000000000000000-ans[i]<<endl;
	}
	return 0;
}
