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>
using namespace std;
typedef long long ll;
typedef pair<int,int> iint;
typedef pair<ll,ll> llll;
#define ALL(x) (x).begin(),(x).end()
const ll zero = 0;
const ll INF = 3000000000000000000; //10^18
const int inINF = 1000000000; //10^9
const ll MOD = 1000000007; //10^9+7
const ll MOD2 = 998244353;



struct edge{
    ll to;
    ll cost;
};

void dijkstra(ll s, vector<vector<edge>> G, vector<ll> &d){
    //first : 最短距離、second : 頂点番号
    priority_queue< llll, vector<llll>, greater<llll>> que;
    fill(ALL(d), INF);

    d[s] = 0;
    que.push(make_pair(0, s));

    while(!que.empty()){
        llll q = que.top(); que.pop();
        ll v = q.second;

        if(q.first > d[v]) continue;

        for (edge e : G[v]) {
            if(q.first + e.cost >= d[e.to]) continue;
            d[e.to] = q.first + e.cost;
            que.push(make_pair(d[e.to], e.to));
        }
    }
}

int main(){
    ll n, m, s, t; cin >> n >> m >> s >> t;
    vector<vector<edge>> Gy(n+1);
    vector<vector<edge>> Gs(n+1);
    vector<ll> ds(n+1);
    vector<ll> dt(n+1);



    ll tmp1, tmp2, tmp3, tmp4;
    for (int i = 0; i < m; i++) {
        cin >> tmp1 >> tmp2 >> tmp3 >> tmp4;
        Gy[tmp1].push_back(edge{tmp2, tmp3});
        Gs[tmp1].push_back(edge{tmp2, tmp4});
        Gy[tmp2].push_back(edge{tmp1, tmp3});
        Gs[tmp2].push_back(edge{tmp1, tmp4});
    }

    dijkstra(s, Gy, ds);
    dijkstra(t, Gs, dt);

    vector<ll> ans(n);
    ans[n-1] = ds[n] + dt[n];
    for (ll i = n-2; i >= 0; i--) {
        ans[i] = min(ans[i+1], ds[i+1] + dt[i+1]);
    }

    ll moto = 1000000000000000;
    for (ll i = 0; i < n; i++) {
        printf("%lld\n", moto - ans[i]);
    }


}