#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,int> P;

const ll INF = 1e18;

int main(){
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--; t--;
  vector<vector<P>> G1(n), G2(n);
  rep(i,m) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--; v--;
    G1[u].push_back({v,a});
    G1[v].push_back({u,a});
    G2[u].push_back({v,b});
    G2[v].push_back({u,b});
  }

  auto dijkstra = [&](int s, vector<vector<P>> &G) {
    vector<ll> d(n,INF);
    priority_queue<pair<ll, int>, vector<pair<ll,int>>, greater<pair<ll,int>>> pq;
    d[s] = 0;
    pq.push({0,s});

    while (!pq.empty()) {
      auto p = pq.top(); pq.pop();
      ll cost = p.first;
      int u = p.second;
      if (d[u] < cost) continue;
      for (auto v : G[u]) {
        if (d[u] + v.second < d[v.first]) {
          d[v.first] = d[u] + v.second;
          pq.push({d[v.first], v.first});
        }
      }
    }
    return d;
  };

  auto d1 = dijkstra(s,G1);
  auto d2 = dijkstra(t,G2);
  vector<ll> res;
  ll tmp = 1e18;
  for (int i = n-1; i >= 0; i--) {
    res.push_back(min(tmp, d1[i] + d2[i]));
    tmp = res.back();
  }
  for (int i = n-1; i >= 0; i--) {
    cout << 1000000000000000 - res[i] << endl;
  }
  return 0;
} 
### Prompt

Please create a solution in cpp to 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;
#define rep(i,n) for(int i = 0; i < (n);i++)
#define sz(x) int(x.size())
typedef long long ll;
typedef pair<int,int> P;

const ll INF = 1e18;

int main(){
  int n, m, s, t;
  cin >> n >> m >> s >> t;
  s--; t--;
  vector<vector<P>> G1(n), G2(n);
  rep(i,m) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--; v--;
    G1[u].push_back({v,a});
    G1[v].push_back({u,a});
    G2[u].push_back({v,b});
    G2[v].push_back({u,b});
  }

  auto dijkstra = [&](int s, vector<vector<P>> &G) {
    vector<ll> d(n,INF);
    priority_queue<pair<ll, int>, vector<pair<ll,int>>, greater<pair<ll,int>>> pq;
    d[s] = 0;
    pq.push({0,s});

    while (!pq.empty()) {
      auto p = pq.top(); pq.pop();
      ll cost = p.first;
      int u = p.second;
      if (d[u] < cost) continue;
      for (auto v : G[u]) {
        if (d[u] + v.second < d[v.first]) {
          d[v.first] = d[u] + v.second;
          pq.push({d[v.first], v.first});
        }
      }
    }
    return d;
  };

  auto d1 = dijkstra(s,G1);
  auto d2 = dijkstra(t,G2);
  vector<ll> res;
  ll tmp = 1e18;
  for (int i = n-1; i >= 0; i--) {
    res.push_back(min(tmp, d1[i] + d2[i]));
    tmp = res.back();
  }
  for (int i = n-1; i >= 0; i--) {
    cout << 1000000000000000 - res[i] << endl;
  }
  return 0;
} 
```