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: Runtime Error)

#include <bits/stdc++.h>
using namespace std;
using ll = long long int;
using namespace std;
#define INF 1000000000000000LL
#define rep(i,a,b) for(int (i)=a;(i)<(int)(b);(i)++)
using ipair = std::pair<ll,ll>;


struct Edge {
    ll post,weight;
    bool operator<(const Edge& o) const{
        return weight < o.weight;
    }
};

ll maxV = 10010;
vector<vector<Edge>>E1(maxV);
vector<vector<Edge>>E2(maxV);
vector<ll> dist1(maxV,1e15);
vector<ll> dist2(maxV,1e15);

void dijkstra1(ll start) {
    priority_queue<ipair, vector<ipair>, greater<ipair>> que;
    dist1[start] = 0;
    que.push(make_pair(0, start));

    while (!que.empty()) {
        ll now_dist,pre;
        tie(now_dist,pre) = que.top();
        que.pop();
        if (dist1[pre] < now_dist) continue;
        for(Edge e:E1[pre]){
            if (dist1[e.post] > dist1[pre] + e.weight) {
                dist1[e.post] = dist1[pre] + e.weight;
                que.push(make_pair(dist1[e.post], e.post));
            }
        }
    }
}
void dijkstra2(ll start) {
    priority_queue<ipair, vector<ipair>, greater<ipair>> que;
    dist2[start] = 0;
    que.push(make_pair(0, start));

    while (!que.empty()) {
        ll now_dist,pre;
        tie(now_dist,pre) = que.top();
        que.pop();
        if (dist2[pre] < now_dist) continue;
        for(Edge e:E2[pre]){
            if (dist2[e.post] > dist2[pre] + e.weight) {
                dist2[e.post] = dist2[pre] + e.weight;
                que.push(make_pair(dist2[e.post], e.post));
            }
        }
    }
}


int main() {
    int n,m,s,t;// Node num, Edge num, start point ,goal point
    cin >> n >> m >> s>> t;
    ll u,v,a,b;
    rep(i,0,m){
        cin >> u >> v >> a >> b;
        u --; v--;
        E1[u].push_back(Edge{v,a});
        E2[u].push_back(Edge{v,b});
        E1[v].push_back(Edge{u,a});
        E2[v].push_back(Edge{u,b});
    }
    ll money = 1e15;
    dijkstra1(s-1);
    dijkstra2(t-1);
    vector<ll> sumDist(n),ans(n);
    rep(i,0,n) sumDist[i] = dist1[i]+dist2[i];

    ll mi = sumDist[n-1];
    for(ll i=n-1;i>=0;i--){
        mi = min(mi,sumDist[i]);
        ans[i] = money - mi;
    }

    for(int i=0;i<n;i++){
        cout << ans[i] << endl;
    }

}