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

#include <cstdio>
#include <vector>
#include <cstring>
#include <string>
#include <queue>
using namespace std;

typedef long long ll;
typedef pair<ll, int> pli;

struct st {
  int to;
  int yen;
  int snu;
};

int main() {
  int n, m, s, t; scanf("%d%d%d%d", &n, &m, &s, &t); s--; t--;
  vector<st> e[n];
  for (int i = 0; i < m; ++i) {
     int u, v, a, b; scanf("%d%d%d%d", &u, &v, &a, &b); u--; v--;
     e[u].push_back((st){v, a, b});
     e[v].push_back((st){u, a, b});
  }

  ll dists[2][n]; memset(dists, 0x3f, sizeof(dists));
  for (int i = 0; i < 2; ++i) {
    ll* dist = dists[i];
    priority_queue<pli> q;
    q.push(make_pair(-0LL, (i == 0 ? s : t)));
    dist[(i == 0 ? s : t)] = 0;
    while(q.size()) {
      ll cost = -q.top().first;
      int pos = q.top().second;
      q.pop();

      for (const st& edge : e[pos]) {
        ll next_cost = cost + (i == 0 ? edge.yen : edge.snu);
        int next_pos = edge.to;
        if (dist[next_pos] <= next_cost) continue;
        dist[next_pos] = next_cost;
        q.push(make_pair(-next_cost, next_pos));
      }
    }
  }

#if 0
  for (int i = 0 ; i < n; ++i) {
    printf("%lld %lld\n", dists[0][i], dists[1][i]);
  }
#endif

  ll ans[n];
  ans[n-1] = dists[0][n-1] + dists[1][n-1];
  for (int i = n - 2; i >= 0; --i) {
    ans[i] = min<ll>(ans[i + 1], dists[0][i] + dists[1][i]);
  }
  for (int i = 0; i < n; ++i) {
    printf("%lld\n", 1000LL * 1000 * 1000 * 1000 * 1000 - ans[i]);
  }
}

