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 (Clang 3.8.0) (judge verdict: Runtime Error)

#include <iostream>
#include <cstdio>
#include <queue>
#include <cmath>
#include <random>

#define REP(i, N) for(int i = 0; i< N; i++)
using namespace std;

int main() {
    std::random_device rnd;     // 非決定的な乱数生成器を生成
    std::mt19937 mt(rnd());     //  メルセンヌ・ツイスタの32ビット版、引数は初期シード値
    std::uniform_int_distribution<> rand100(1, 100000);        // [0, 99] 範囲の一様乱数
    std::uniform_int_distribution<> rand1000(1, 1000000000);
    
    for (int i = 0; i < 20; ++i) {
        std::cout << rand100(mt) << "\n";
    }
    long long int n = rand100(mt);
    long long int m = rand100(mt) % (n * (n - 1) / 2);
    std::uniform_int_distribution<> randn(1, n);
    
    
    long long int s = randn(mt);
    long long int t = s;
    while(t == s){t = randn(mt);}
    int es, et;
    
    
    long long int graph[100000][100000] = {0};
    REP(i, m){
        do{
            es = rnd() % n;
            et = rnd() % (n - es) + es;
        }while(graph[es][et] == 1);
        graph[es][et] = 1;
        printf("%lld %lld %lld %lld\n", es, et, rand1000(mt), rand1000(mt));
    }
    }