#include <bits/stdc++.h>
using namespace std;
#define rep(i, m, n) for (int i = m; i < n; ++i)
#define rem(i, m, n) for (int i = m; i >= n; --i)
typedef long long ll;

typedef pair<ll, int> P;
const ll INF = 1e15;
struct edge {
  int to;
  ll cost;
};

class Dijk {
public:
  vector<edge> G[100005];
  ll dist[100005];
  int used[100005];

  void dijk(int s, int V) {
    priority_queue<P, vector<P>, greater<P>> Q;
    rep(i, 0, V) dist[i] = INF;
    rep(i, 0, V) used[i] = 0;
    dist[s] = 0;
    Q.push({0, s});

    while (!Q.empty()) {
      P p = Q.top();
      Q.pop();
      int v = p.second;
      if (used[v]) continue;
      used[v] = 1;
      for (auto &e : G[v]) {
        if (dist[e.to] > dist[v] + e.cost) {
          dist[e.to] = dist[v] + e.cost;
          Q.push({dist[e.to], e.to});
        }
      }
    }
  }
};

int main() {
  int N, M, s, t;
  cin >> N >> M >> s >> t;
  s--, t--;
  Dijk D1, D2;
  rep(i, 0, M) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--, v--;
    D1.G[u].push_back({v, a});
    D1.G[v].push_back({u, a});
    D2.G[u].push_back({v, b});
    D2.G[v].push_back({u, b});
  }
  D1.dijk(s, N);
  D2.dijk(t, N);
  vector<ll> ans(N);
  rep(i, 0, N) ans[i] = D1.dist[i] + D2.dist[i];
  rem(i, N-1, 1) ans[i-1] = min(ans[i], ans[i-1]);
  rep(i, 0, N) printf("%lld\n", (ll)1e15 - ans[i]);
  return 0;
}

### Prompt

Develop a solution in Cpp to the problem described below:

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, m, n) for (int i = m; i < n; ++i)
#define rem(i, m, n) for (int i = m; i >= n; --i)
typedef long long ll;

typedef pair<ll, int> P;
const ll INF = 1e15;
struct edge {
  int to;
  ll cost;
};

class Dijk {
public:
  vector<edge> G[100005];
  ll dist[100005];
  int used[100005];

  void dijk(int s, int V) {
    priority_queue<P, vector<P>, greater<P>> Q;
    rep(i, 0, V) dist[i] = INF;
    rep(i, 0, V) used[i] = 0;
    dist[s] = 0;
    Q.push({0, s});

    while (!Q.empty()) {
      P p = Q.top();
      Q.pop();
      int v = p.second;
      if (used[v]) continue;
      used[v] = 1;
      for (auto &e : G[v]) {
        if (dist[e.to] > dist[v] + e.cost) {
          dist[e.to] = dist[v] + e.cost;
          Q.push({dist[e.to], e.to});
        }
      }
    }
  }
};

int main() {
  int N, M, s, t;
  cin >> N >> M >> s >> t;
  s--, t--;
  Dijk D1, D2;
  rep(i, 0, M) {
    int u, v, a, b;
    cin >> u >> v >> a >> b;
    u--, v--;
    D1.G[u].push_back({v, a});
    D1.G[v].push_back({u, a});
    D2.G[u].push_back({v, b});
    D2.G[v].push_back({u, b});
  }
  D1.dijk(s, N);
  D2.dijk(t, N);
  vector<ll> ans(N);
  rep(i, 0, N) ans[i] = D1.dist[i] + D2.dist[i];
  rem(i, N-1, 1) ans[i-1] = min(ans[i], ans[i-1]);
  rep(i, 0, N) printf("%lld\n", (ll)1e15 - ans[i]);
  return 0;
}

```