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;

vector<pair<int, long long>> Y[100010];
vector<pair<int, long long>> S[100010];

bool vy[100010];
bool vs[100010];

long long ny[100010],ns[100010];

priority_queue<pair<long long, int>> H;

int main()
{
    int n,m,s,t;
    int a,b,y,sn;
    cin >> n >> m >> s >> t;
    for ( int c=1; c<=m; c++ )
    {
        cin >> a >> b >> y >> sn;
        Y[a].push_back({b,y});
        Y[b].push_back({a,y});
        S[a].push_back({b,sn});
        S[b].push_back({a,sn});
    }

    H.push({0,s});
    while(!H.empty())
    {
        auto pa = H.top();
        H.pop();

        if (vy[pa.second])
            continue;
        vy[pa.second] = true;
        ny[pa.second] = -pa.first;
        for ( auto ed: Y[pa.second])
        {
            H.push({pa.first-ed.second, ed.first});
        }
    }

    H.push({0,t});
    while(!H.empty())
    {
        auto pa = H.top();
        H.pop();

        if (vs[pa.second])
            continue;
        vs[pa.second] = true;
        ns[pa.second] = -pa.first;
        for ( auto ed: S[pa.second])
        {
            H.push({pa.first-ed.second, ed.first});
        }
    }


    long long mx=1e15;
    vector<long long> ans(100010);

    for ( int c=n; c>=1; c-- )
    {
        mx = min (mx, ny[c]+ns[c]);
        ans[c] = 1e15 - mx;
    }

    for ( int c=1; c<=n; c++ )
    {
        cout << ans[c] << endl;
    }

}
