#include<bits/stdc++.h>
using namespace std;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long

struct edge{ int to,cost; };
vector<int> dikstra(int s,vector<vector<edge>> G){
  typedef pair<int, int> P;
  const int INF = 1LL << 60;

  priority_queue<P,vector<P>,greater<P>> que;
  vector<int> d(G.size(),INF); //sからの最短距離
  d[s] = 0;
  que.push({0,s}); //{最短距離,頂点}

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

signed main(){

  int n, m, s, t; cin >> n >> m >> s >> t;
  s--, t--;
  vector<vector<edge>> yen(n), snu(n);
  for(int i = 0; i < m; i++){
    int u, v, a, b; cin >> u >> v >> a >> b;
    u--, v--;
    yen[u].push_back({v, a});
    yen[v].push_back({u, a});
    snu[u].push_back({v, b});
    snu[v].push_back({u, b});
  }

  auto dp = dikstra(s, yen);
  auto dp2 = dikstra(t, snu);

  vector<int> ans(n);
  int res = 1LL << 60;
  for(int i = n-1; i >= 0; i--){
    res = min(res, dp[i] + dp2[i]);
    ans[i] = res;
  }

  for(int i = 0; i < n; i++){
    cout << (int)1e15 - ans[i] << endl;
  }

  return 0;
}

### Prompt

Generate a CPP solution 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;
#ifdef LOCAL_DEBUG
  #include "LOCAL_DEBUG.hpp"
#endif
#define int long long

struct edge{ int to,cost; };
vector<int> dikstra(int s,vector<vector<edge>> G){
  typedef pair<int, int> P;
  const int INF = 1LL << 60;

  priority_queue<P,vector<P>,greater<P>> que;
  vector<int> d(G.size(),INF); //sからの最短距離
  d[s] = 0;
  que.push({0,s}); //{最短距離,頂点}

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

signed main(){

  int n, m, s, t; cin >> n >> m >> s >> t;
  s--, t--;
  vector<vector<edge>> yen(n), snu(n);
  for(int i = 0; i < m; i++){
    int u, v, a, b; cin >> u >> v >> a >> b;
    u--, v--;
    yen[u].push_back({v, a});
    yen[v].push_back({u, a});
    snu[u].push_back({v, b});
    snu[v].push_back({u, b});
  }

  auto dp = dikstra(s, yen);
  auto dp2 = dikstra(t, snu);

  vector<int> ans(n);
  int res = 1LL << 60;
  for(int i = n-1; i >= 0; i--){
    res = min(res, dp[i] + dp2[i]);
    ans[i] = res;
  }

  for(int i = 0; i < n; i++){
    cout << (int)1e15 - ans[i] << endl;
  }

  return 0;
}

```