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<algorithm>
#include<queue>
#include<vector>
#include<utility>
#include<stdlib.h>
#include<string.h>
using namespace std;

#define se second
#define fi first
#define mk(a,b) make_pair(a,b)

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

int N,M,S,T;
vector<pii> G[2][100010];
ll D1[100010],D2[100010];
ll ans = 0;
vector<ll> mem;

void daijkstra(ll s,ll *d,int k)
{
    priority_queue<pii,vector<pii>,greater<pii> > q;
    q.push(mk(0,s));
    for(int i = 0; i <= N; i++)d[i] = -1;
    while(q.size())
    {
        pii now = q.top();
        q.pop();
        if(d[now.se] != -1)continue;
        d[now.se] = now.fi;
        for(int i = 0; i < G[k][now.se].size(); i++)
            q.push(mk(now.fi + G[k][now.se][i].fi,G[k][now.se][i].se));
    }
    return;
}

int main()
{
    scanf("%d%d%d%d",&N,&M,&S,&T);

    for(int i = 0; i < M; i++)
    {
        int u,v;
        ll a,b;
        scanf("%d%d%lld%lld",&u,&v,&a,&b);
        G[0][u].push_back(mk(a,v));
        G[0][v].push_back(mk(a,u));
        G[1][u].push_back(mk(b,v));
        G[1][v].push_back(mk(b,u));
    }
    daijkstra(S,D1,0);
    daijkstra(T,D2,1);

    for(int i = N; i >= 1; i--)
    {
        ans = max(ans,1000000000000000 - D1[i] - D2[i]);
        mem.push_back(ans);
    }
    for(int i = N - 1; i >= 0; i--)printf("%lld\n",mem[i]);

    return 0;
}
