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: Accepted)

#include <bits/stdc++.h>
#include <math.h>
using namespace std;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)
#define rep2(i, s, n) for (int i = (s); i < (int)(n); i++)

const int MOD = 1000000007;
const int INF_32 = 1LL << 30;
const int64_t INF_64 = 1LL << 60;

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;
}

typedef long long ll;

#define MAX_N 10000
#define ORD_MONEY 1000000000000000

struct Edge {
    int to;
    ll weight_yen;
    ll weight_sunuke;
    Edge(int t, ll y, ll s)
        : to(t)
        , weight_yen(y)
        , weight_sunuke(s)
    {
    }
};
using Graph = vector<vector<int>>;
using Graph_e = vector<vector<Edge>>;

vector<int> dx = { 1, 0, -1, 0 };
vector<int> dy = { 0, 1, 0, -1 };
typedef pair<ll, int> P; // first:最短距離，second:頂点番号

void dijkstra(int s, int V, Graph_e G, ll* d, bool mode)
{ // mode=trueならsから，falseならtから
    fill(d, d + V, INF_64);
    d[s] = 0;

    priority_queue<P, vector<P>, greater<P>> que; // 距離が小さいものから取り出せるように
    que.push(P(0, s));

    while (!que.empty()) {
        P p = que.top();
        que.pop();
        int v = p.second; //更新した頂点の中で距離が最小の頂点
        if (d[v] < p.first) {
            continue;
        }

        for (auto e : G[v]) {
            if (mode) {
                if (d[e.to] > d[v] + e.weight_yen) {
                    d[e.to] = d[v] + e.weight_yen;
                    que.push(P(d[e.to], e.to));
                }
            } else {
                if (d[e.to] > d[v] + e.weight_sunuke) {
                    d[e.to] = d[v] + e.weight_sunuke;
                    que.push(P(d[e.to], e.to));
                }
            }
        }
    }
}

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

    Graph_e G(n);
    ll d_s[n];
    ll d_t[n];

    rep(i, m)
    {
        int u, v, a, b;
        cin >> u >> v >> a >> b;
        u--;
        v--;
        G[u].push_back(Edge(v, a, b));
        G[v].push_back(Edge(u, a, b));
    }
    // どこかで両替したら以降はずっとsunuke
    //
    // 両替を都市iで行う場合，
    // sからiに円による支払いで移動する際の最小金額（yen)
    // iからtにsunukeによる支払いで移動する際の最小金額（sunuke）
    // の和になる
    // tからiにスヌークによる支払いで移動する際の最小金額とiからtにスヌークによる支払いで移動する際の最小金額は同じ
    // これらをダイクストラ法でそれぞれ求めれば良い

    dijkstra(s, n, G, d_s, true);
    dijkstra(t, n, G, d_t, false);

    ll d_sum[n];
    ll ans[n];
    rep(i, n)
    {
        d_sum[i] = d_s[i] + d_t[i];
    }

    ll min = INF_64;
    for (int i = n - 1; i >= 0; i--) {
        if (d_sum[i] < min) {
            min = d_sum[i];
        }
        ans[i] = ORD_MONEY - min;
    }

    rep(i, n)
    {
        cout << ans[i] << endl;
    }
}