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 all(a) (a).begin(), (a).end()
#define rall(a) (a).rbegin(), (a).rend()
#define MIN(a) *min_element(all(a))
#define MAX(a) *max_element(all(a))
#define SUM(a, b) accumulate(all(a), b)
#define REP(i, n) for(int (i)=0; (i)<(n); (i)++)
#define RREP(i, n) for(int (i)=(n)-1; (i)>=0; (i)--)
#define FOR(i, m, n) for(int (i)=(m); (i)<(n); i++)
#define FORR(i, m, n) for(int (i)=(n)-1; (i)>=(m); i--)
#define ios() cin.tie(0), ios::sync_with_stdio(false)
#define debug(x) cerr << #x << " = " << x << endl
//#define int long long

typedef long long ll;
typedef unsigned long long ull;
//ll const MOD = 998244353;
ll const MOD = 1e9+7;
ll const INF = 1e18;

template<typename T> int size(const T& a){return (int)a.size();}
template<typename T> T Div(T a, T b){return (a + b - 1) / b;}
template<typename T> bool chmin(T& a, const T& b){if(a > b){a = b; return true;} return false;}
template<typename T> bool chmax(T& a, const T& b){if(a < b){a = b; return true;} return false;}

//Graph in adjcency list (weighted)
class Graph{
public:
    int N;
    vector<vector<pair<int,ll>>> adjlist;
    vector<ll> depth;

    Graph(int n) : N(n), adjlist(n), depth(n, -1){}

    void setEdge(int a, int b, ll c){
        adjlist[a].push_back(make_pair(b, c));
    }

    void dfs(int p, int v, ll dist){
        static vector<bool> check(N, false);
        if(check[v]){
            return;
        }
        check[v] = true;
        depth[v] = dist;
        for(auto i : adjlist[v]){
            if(i.first == p){
                continue;
            }
            dfs(v, i.first, dist + i.second);
        }
    }

    void dijkstra(int start){
        depth[start] = 0;

        auto fn = [](pair<int,ll> a, pair<int, ll> b){return a.second > b.second;};
        priority_queue<pair<int,ll>, vector<pair<int,ll>>, decltype(fn)> que(fn);

        for(auto i : adjlist[start]) que.push(i);

        while(!que.empty()){
            auto v = que.top().first;
            auto d = que.top().second;
            que.pop();

            if(depth[v] >= 0) continue;
            else depth[v] = d;
            
            for(auto i : adjlist[v]){
                if(depth[i.first] == -1){
                    que.push(make_pair(i.first, i.second + d));
                }
            }
        }
    }
};


signed main(){
    int n, m, start, goal;
    cin >> n >> m >> start >> goal;
    start--;
    goal--;

    Graph GY(n), GS(n);

    REP(i, m){
        int u, v;
        ll a, b;
        cin >> u >> v >> a >> b;
        u--;
        v--;
        GY.setEdge(u, v, a);
        GY.setEdge(v, u, a);
        GS.setEdge(u, v, b);
        GS.setEdge(v, u, b);
    }

    GY.dijkstra(start);
    GS.dijkstra(goal);

    vector<ll> cost(n+1);
    cost[n] = INF;

    RREP(i, n){
        cost[i] = min(cost[i+1], GY.depth[i] + GS.depth[i]);
    }

    REP(i, n){
        cout << (ll)1e15 - cost[i] << '\n';
    }

    return 0;
}