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<iostream>
#include<cstdio>
#include<vector>
#include<queue>
typedef long long ll;
typedef std::pair<ll,int> P;
const int N=1e5+5;
const ll INF=1e18;
ll tot=1e15;
int n,m,s,t;
std::vector<int> g[N],e[2][N];
ll d[2][N],ans[N];
void dijkstra(int s,ll *dis,const std::vector<int> *e) {
    for(int i=1;i<=n;i++) dis[i]=INF;dis[s]=0;
    std::priority_queue<P,std::vector<P>,std::greater<P> > q;
    q.push(P(0,s));
    while(!q.empty()) {
        P u=q.top();q.pop();
        if(dis[u.second]<u.first) continue;
        for(int i=0;i<g[u.second].size();i++) {
            int v=g[u.second][i],w=e[u.second][i];
            if(dis[v]>dis[u.second]+w) {
                dis[v]=dis[u.second]+w;
                q.push(P(dis[v],v));
            }
        }
    }
}
int main() {
    scanf("%d%d%d%d",&n,&m,&s,&t);
    int u,v,a,b;
    while(m--) {
        scanf("%d%d%d%d",&u,&v,&a,&b);
        g[u].push_back(v);e[0][u].push_back(a);e[1][u].push_back(b);
        g[v].push_back(u);e[0][v].push_back(a);e[1][v].push_back(b);
    }
    dijkstra(s,d[0],e[0]);dijkstra(t,d[1],e[1]);
    ans[n]=0;
    for(int i=n-1;~i;i--) ans[i]=std::max(ans[i+1],tot-d[0][i+1]-d[1][i+1]);
    for(int i=0;i<n;i++) printf("%lld\n",ans[i]);
    return 0;
}
