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>

#define f first
#define s second
#define pb push_back

using namespace std;

typedef long long ll;
typedef pair<ll, ll> pll;
typedef pair<ll, int> pli;
typedef pair<int, int> pii;
typedef pair<int, pll> pill;

const int maxk = 17 + 1;
const int maxn = 1 * 1e5 + 17;

int N, M, S, T;
ll DPS[maxn], DPT[maxn];

vector<pill> graph[maxn];

void solve(int from, ll *DP)
{
    for (int i = 0; i < maxn; i++) DP[i] = 1LL << 62;
    priority_queue<pll, vector<pll>, greater<pll> > pq;
    DP[from] = 0; pq.push({DP[from], from});

    while (!pq.empty()) {
        int u = pq.top().s;
        pq.pop();

        for (auto e : graph[u]) {
            ll len = e.s.f;
            if (DP == DPT) len = e.s.s;

            if (DP[u] + len < DP[e.f])
                DP[e.f] = DP[u] + len,
                pq.push({DP[e.f], e.f});
        }
    }
}

int main()
{
    ios_base::sync_with_stdio(false);

    cin >> N >> M >> S >> T;
    for (int i = 1; i <= M; i++) {
        int x, y; ll z, w; cin >> x >> y >> z >> w;
        graph[x].pb({y, {z, w}}); graph[y].pb({x, {z, w}});
    }

    solve(S, DPS); solve(T, DPT);

    priority_queue<pli, vector<pli>, greater<pli> > ans;
    for (int i = 1; i <= N; i++) ans.push({DPS[i] + DPT[i], i});

    for (int i = 0; i <= N - 1; i++) {
        while (ans.top().s <= i) ans.pop();
        cout << 1000000000000000LL - ans.top().f << endl;
    }

    return 0;
}