#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> P;
int n, m, s, t;
vector<P> graph[100001];

vector<long long> dijkstra(int start, vector<int> costs) {
  priority_queue<P, vector<P>, greater<P> > que;
  vector<long long> d(n);
  for (int i=0; i<n; i++) d[i] = 1e14;

  d[start] = 0;
  que.push(P(0, start));

  while (!que.empty()) {
    P p = que.top();
    que.pop();
    int v = p.second;
    if (d[v] < p.first) continue;
    for (int i=0; i<graph[v].size(); i++) {
      auto e = graph[v][i];
      if (d[e.first] > d[v] + costs[e.second]) {
        d[e.first] = d[v] + costs[e.second];
        que.push(P(d[e.first], e.first));
      }
    }
  }
  return d;
}
int main() {

  cin>>n>>m>>s>>t;
  --s;
  --t;
  vector<int> a(m), b(m);
  for (int i=0; i<m; i++) {
    int ui, vi;
    cin>>ui>>vi>>a[i]>>b[i];
    --ui;
    --vi;
    graph[ui].push_back(P(vi, i));
    graph[vi].push_back(P(ui, i));
  }
  vector<long long> d_yen = dijkstra(s, a);
  vector<long long> d_snk = dijkstra(t, b);

  vector<long long> ans;
  long long init = 1e15;
  ans.push_back(1e15 - d_yen[n-1] - d_snk[n-1]);
  for (int i=1; i<n; i++) {
    long long tmp = 1e15 - d_yen[n-i-1] - d_snk[n-i-1];
    ans.push_back(max(tmp, ans[i-1]));
  }
  reverse(ans.begin(), ans.end());
  for (auto a: ans) {
    cout<<a<<endl;
  }
}
### Prompt

In Cpp, your task is to solve the following problem:

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.

Examples

Input

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1


Output

999999999999998
999999999999989
999999999999979
999999999999897


Input

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


Output

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef pair<int, int> P;
int n, m, s, t;
vector<P> graph[100001];

vector<long long> dijkstra(int start, vector<int> costs) {
  priority_queue<P, vector<P>, greater<P> > que;
  vector<long long> d(n);
  for (int i=0; i<n; i++) d[i] = 1e14;

  d[start] = 0;
  que.push(P(0, start));

  while (!que.empty()) {
    P p = que.top();
    que.pop();
    int v = p.second;
    if (d[v] < p.first) continue;
    for (int i=0; i<graph[v].size(); i++) {
      auto e = graph[v][i];
      if (d[e.first] > d[v] + costs[e.second]) {
        d[e.first] = d[v] + costs[e.second];
        que.push(P(d[e.first], e.first));
      }
    }
  }
  return d;
}
int main() {

  cin>>n>>m>>s>>t;
  --s;
  --t;
  vector<int> a(m), b(m);
  for (int i=0; i<m; i++) {
    int ui, vi;
    cin>>ui>>vi>>a[i]>>b[i];
    --ui;
    --vi;
    graph[ui].push_back(P(vi, i));
    graph[vi].push_back(P(ui, i));
  }
  vector<long long> d_yen = dijkstra(s, a);
  vector<long long> d_snk = dijkstra(t, b);

  vector<long long> ans;
  long long init = 1e15;
  ans.push_back(1e15 - d_yen[n-1] - d_snk[n-1]);
  for (int i=1; i<n; i++) {
    long long tmp = 1e15 - d_yen[n-i-1] - d_snk[n-i-1];
    ans.push_back(max(tmp, ans[i-1]));
  }
  reverse(ans.begin(), ans.end());
  for (auto a: ans) {
    cout<<a<<endl;
  }
}
```