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: Wrong Answer)

#include<cstdio>
#include<queue>
#include<utility>
#include<cstring>
#include<stack>
#include<algorithm>
#include<cmath>
#include<iostream>
#include<vector>
using namespace std;
#define MAX_N 100001
#define INF 2147483647
#define REP(i,COUNT) for(int i=0;i<(int)(COUNT);i++)
struct edge{
  long long int to, cost;
};
typedef vector<vector<edge> > AdjList;
AdjList graph;
AdjList graph2;
typedef pair<long long int, long long int> P;
void init(int n);
int find(int n);
void unite(int x,int y);
bool same(int x, int y);
int dx[4] = {1,0,0,-1};
int dy[4] = {0,1,-1,0};
vector<long long int> dist;
vector<long long int> prever;
void dijkstra(int n,long long int s){
  dist = vector<long long int>(n,INF);
  prever = vector<long long int>(n,-1);
  dist[s] = 0;

  priority_queue<P, vector<P> , greater<P> > que;
  que.push(P(0,s));

  while(!que.empty()){
    P p = que.top();
    que.pop();
    int v = p.second;
    if(dist[v] < p.first)
      continue;
    REP(i,graph[v].size()){
      edge e = graph[v][i];
      if(dist[e.to] > dist[v] + e.cost){
        dist[e.to] = dist[v] + e.cost;
        prever[e.to] = v;
        que.push(P(dist[e.to],e.to));
      }

    }

  }
}

int main()
{
  long long int money = 1.0E15,res[100001];
  long long int u,v,n,m,s,t,a,b,distt[100001];
  cin >> n >> m >> s >> t;
  graph = AdjList(n);
  graph2 = AdjList(n);
  REP(i,m){
    cin >> u >> v  >> a >> b;
    graph[u-1].push_back((edge){v-1,a});
    graph[v-1].push_back((edge){u-1,a});
    graph2[u-1].push_back((edge){v-1,b});
    graph2[v-1].push_back((edge){u-1,b});
  }
  
  dijkstra(n,s-1);
  REP(i,n)distt[i] = dist[i];
  graph = graph2;
  dijkstra(n,t-1);
  res[n-1] = distt[n-1] + dist[n-1];
  for(int i=n-2;i>=0;i--)
    res[i] = min(distt[i] + dist[i],res[i+1]);
  REP(i,n)
    cout << money - res[i] << endl;
  return 0;
}

int par[MAX_N];
int ranks[MAX_N];

//n要素で初期化
void init(int n){
  REP(i,n){
    par[i] = i;
    ranks[i] = 0;
  }

}

//木の根を求める
int find(int x){
  if(par[x] == x){
    return x;
  }else{
    return par[x] = find(par[x]);
  }
}

void unite(int x,int y){
  x = find(x);
  y = find(y);
  if(x == y) return ;
  if(ranks[x] < ranks[y]){
    par[x] = y;
  }else{
    par[y] = x;
    if(ranks[x] == ranks[y]) ranks[x]++;
  }
}

bool same(int x, int y){
  return find(x) == find(y);
}