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;
typedef double db;
typedef pair<ll,ll> P;
#define pb push_back
#define mp make_pair
#define fr(i,n) for(int i=0;i<n;i++)
#define Fr(i,n) for(int i=0;i++<n;)
#define ifr(i,n) for(int i=n-1;i>=0;i--)
#define iFr(i,n) for(int i=n;i>0;i--)

ll n,m;

void dkstr(vector<vector<P>>& e,ll s,vector<ll>& result){
  result.resize(0);result.resize(n+1);
  priority_queue<P,vector<P>,greater<P>> pq;
  vector<bool> vis(n+1);
  pq.push(mp(0,s));
  while(!pq.empty()){
    ll i=pq.top().second,c=pq.top().first;pq.pop();
    if(vis[i]) continue;
    vis[i]=1;result[i]=c;
    fr(j,e[i].size()) if(!vis[e[i][j].first]) pq.push(mp(c+e[i][j].second,e[i][j].first));
  }
}

int main(){
  ll s,t,u,v,a,b,mn[100010];
  vector<ll> ra,rb;
  cin>>n>>m>>s>>t;
  vector<vector<P>> ea(n+1),eb(n+1);
  fr(i,m){
    cin>>u>>v>>a>>b;
    ea[u].pb(mp(v,a));
    ea[v].pb(mp(u,a));
    eb[u].pb(mp(v,b));
    eb[v].pb(mp(u,b));
  }
  dkstr(ea,s,ra);dkstr(eb,t,rb);
  mn[n+1]=1e15;
  iFr(i,n) mn[i]=min(ra[i]+rb[i],mn[i+1]);
  Fr(i,n) cout<<1000000000000000-mn[i]<<endl;
}
