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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll N=1e5+5;
ll inf=1e17,MX=1e15;
vector< pair<ll,ll> > g1[N],g2[N];
ll dist1[N],dist2[N],ans[N];
void dijkstra(vector< pair<ll,ll> > g[],ll dist[],ll s,ll n){
    for(ll i=1;i<=n;i++)dist[i]=inf;
    dist[s]=0;
    priority_queue< pair<ll,ll> , vector< pair<ll,ll> >  , greater< pair<ll,ll> >  >pq;
    pq.push({0,s});
    while(!pq.empty()){
        auto node=pq.top();
        pq.pop();
        ll u = node.second;
        if(dist[u]<node.first)continue;
        for(auto &v:g[u]){
            if(dist[u]+v.second < dist[v.first]){
                dist[v.first] = dist[u] + v.second;
                pq.push({dist[v.first] ,v.first});
            }
        }
    }
}
int main(){
    ios_base:: sync_with_stdio(false);
    cin.tie(NULL); cout.tie(NULL);
    ll n,m,s,t;cin>>n>>m>>s>>t;
    for(ll i=1;i<=m;i++){
        ll u,v,a,b;cin>>u>>v>>a>>b;
        g1[u].push_back({v,a});
        g1[v].push_back({u,a});
        g2[u].push_back({v,b});
        g2[v].push_back({u,b});
    }
    dijkstra(g1,dist1,s,n);
    dijkstra(g2,dist2,t,n);
    for(ll u=1;u<=n;u++)ans[u]=dist1[u]+dist2[u];
    for(ll i=n-1;i>=1;i--)ans[i]=min(ans[i],ans[i+1]);
    for(ll i=1;i<=n;i++)cout<<MX-ans[i]<<"\n";
}

