Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

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

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define repd(i, a, b) for (int i = (a); i < (b); i++)
#define rep(i, n) repd(i, 0, n)
#define INF 2e9
#define ALL(v) v.begin(), v.end()
using namespace std;
typedef long long ll;
template <class T>
inline bool chmin(T &a, T b)
{
    if (a > b)
    {
        a = b;
        return true;
    }
    return false;
}
template <class T>
inline bool chmax(T &a, T b)
{
    if (a < b)
    {
        a = b;
        return true;
    }
    return false;
}

struct edge
{
    int to;
    int cost;
};

typedef pair<int, int> P;

int main()
{
    int n, m, s, t;
    cin >> n >> m >> s >> t;

    vector<edge> G1[n], G2[n];
    priority_queue<P, vector<P>, greater<P>> que1, que2;
    vector<ll> d1(n, INF), d2(n, INF);

    d1[s - 1] = 0;
    que1.push(P(0, s - 1));
    d2[t - 1] = 0;
    que2.push(P(0, t - 1));

    rep(i, m)
    {
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        edge e1, e2;
        e1.to = v - 1;
        e1.cost = a;
        e2.to = v - 1;
        e2.cost = b;
        G1[u - 1].push_back(e1);
        e1.to = u - 1;
        G1[v - 1].push_back(e1);
        G2[u - 1].push_back(e2);
        e2.to = u - 1;
        G2[v - 1].push_back(e2);
    }

    while (!que1.empty())
    {
        P p = que1.top();
        que1.pop();
        int v = p.second;
        if (d1[v] < p.first)
        {
            continue;
        }
        rep(i, G1[v].size())
        {
            edge e = G1[v][i];
            if (d1[e.to] > d1[v] + e.cost)
            {
                d1[e.to] = d1[v] + e.cost;
                que1.push(P(d1[e.to], e.to));
            }
        }
    }

    while (!que2.empty())
    {
        P p = que2.top();
        que2.pop();
        int v = p.second;
        if (d2[v] < p.first)
        {
            continue;
        }
        rep(i, G2[v].size())
        {
            edge e = G2[v][i];
            if (d2[e.to] > d2[v] + e.cost)
            {
                d2[e.to] = d2[v] + e.cost;
                que2.push(P(d2[e.to], e.to));
            }
        }
    }

    vector<ll> ans(n, INF);

    rep(i, n)
    {
        rep(j, n)
        {
            if (i <= j)
            {
                ans[i] = min(ans[i], d1[j] + d2[j]);
            }
        }
    }

    ll money = pow(10, 15);

    rep(i, n)
    {
        ll res = money - ans[i];
        cout << res << endl;
    }

    return 0;
}