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;

#define lint long long
#define P pair<int, int>
#define LLP pair<long long, long long>
#define REP(i, x, n) for(int i = (x), i##_len = (int)(n) ; i < i##_len ; ++i)
#define rep(i, n) for(int i = 0, i##_len = (int)(n) ; i < i##_len ; ++i)
#define reps(i, n) for(int i = 1, i##_len = (int)(n) ; i <= i##_len ; ++i)
#define rrep(i, n) for(int i = (int)(n) - 1 ; i >= 0 ; --i)
#define rreps(i, n) for(int i = (int)(n) ; i > 0 ; --i)
#define SORT(x) sort((x).begin(), (x).end())
#define SORT_INV(x) sort((x).rbegin(), (x).rend())

const int IINF = (1 << 30) - 1;
const long long LLINF = 1LL << 61;
const long long MOD = 1000000007LL;
const int dx4[] = {1, 0, -1, 0}, dy4[] = {0, 1, 0, -1};
const int dx8[] = {1, 1, 0, -1, -1, -1, 0, 1}, dy8[] = {0, -1, -1, -1, 0, 1, 1, 1};
const double EPS = 1e-8;

struct edge{
    int to;
    lint yen, snuuk;
};

struct node{
    int id;
    lint cost;

    bool operator<(const node & right) const {
        return cost > right.cost;
    }
};

template<typename T>
bool chmin(T &_a, T _b){
    if(_a > _b){
        _a = _b;
        return true;
    }else{
        return false;
    }
}

template<typename T>
bool chmax(T &_a, T _b){
    if(_a < _b){
        _a = _b;
        return true;
    }else{
        return false;
    }
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);

    int n, m, s, t;
    cin >> n >> m >> s >> t;
    --s;
    --t;

    vector< vector<edge> > g(n);
    rep(i, m){
        int u, v;
        lint a, b;
        cin >> u >> v >> a >> b;
        --u;
        --v;
        g[u].push_back({v, a, b});
        g[v].push_back({u, a, b});
    }

    vector<lint> yen(n, LLINF), snuuk(n, LLINF);
    priority_queue<node> que;
    que.push({s, 0LL});
    while(!que.empty()){
        node now = que.top();
        que.pop();
        if(!chmin(yen[now.id], now.cost)){
            continue;
        }
        for(auto e : g[now.id]){
            if(yen[e.to] > now.cost + e.yen){
                que.push({e.to, now.cost + e.yen});
            }
        }
    }
    que.push({t, 0LL});
    while(!que.empty()){
        node now = que.top();
        que.pop();
        if(!chmin(snuuk[now.id], now.cost)){
            continue;
        }
        for(auto e : g[now.id]){
            if(snuuk[e.to] > now.cost + e.snuuk){
                que.push({e.to, now.cost + e.snuuk});
            }
        }
    }

    vector<lint> ans(n, (lint)1e15);
    rep(i, n){
        ans[i] -= yen[i] + snuuk[i];
    }
    rrep(i, n - 1){
        chmax(ans[i], ans[i + 1]);
    }

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

    return 0;
}