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 <stdio.h>
#include <vector>
#include <utility>
#include <tuple>
#include <queue>
using namespace std;
typedef long long ll;
typedef pair<ll, int> P;    // <dist, v>

const int MAX_N = 100001;
const ll INF = 1e18;

struct edge { int to, cost; };
vector<edge> G[2][MAX_N];       // [A/B][N]
ll dist[2][MAX_N];              // [from S/from T]
ll ans[MAX_N];

void dijkstra(vector<edge>* g, ll* d, int s) {
    priority_queue<P, vector<P>, greater<P>> que;
    fill(d, d+MAX_N, INF);
    d[s] = 0;
    que.push({0, s});

    int cost, v;
    while (!que.empty()) {
        tie(cost, v) = que.top();
        que.pop();
        if (d[v] < cost) continue;
        for (const edge& 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});
            }
        }
    }
}

int main() {
    int n, m, s, t;
    int u, v, a, b;
    scanf("%d %d %d %d", &n, &m, &s, &t);
    for (int i = 0; i < m; i++) {
        scanf("%d %d %d %d", &u, &v, &a, &b);
        G[0][u].push_back({v, a});
        G[0][v].push_back({u, a});
        G[1][u].push_back({v, b});
        G[1][v].push_back({u, b});
    }
    dijkstra(G[0], dist[0], s);
    dijkstra(G[1], dist[1], t);
    for (int i = 1; i <= n; i++)
        ans[i-1] = dist[0][i] + dist[1][i];
    for (int i = n-2; i >= 0; i--)
        ans[i] = min(ans[i], ans[i+1]);
    for (int i = 0; i < n; i++)
        printf("%lld\n", (ll)1e15 - ans[i]);
}
