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
/*
By Tianyi Chen. All rights reserved.
Date: 2018-07-07
*/
#include<bits/stdc++.h>
#include<queue>
using namespace std;
int n,m,s,t;
multiset<int64_t>ms;
deque<multiset<int64_t>::iterator>q;
int64_t ds[100001],dt[100001];
vector<pair<int,int64_t>>es[100001],et[100001];
void sh(vector<pair<int,int64_t>>*e,int s,int64_t*d){
 memset(d,0x33,100001*8);
 d[s]=0;
 priority_queue<pair<int64_t,int>,vector<pair<int64_t,int>>,greater<pair<int64_t,int>>>q;
 q.emplace(0,s);
 while(q.size()){
  int p;int64_t dd;
  tie(dd,p)=q.top();q.pop();
  if(dd!=d[p])continue;
  for(auto&&x:e[p])if(d[x.first]>dd+x.second){
   q.emplace(d[x.first]=dd+x.second,x.first);
  }
 }
}
int main(){
 scanf("%d%d%d%d",&n,&m,&s,&t);
 while(m--){
  int u,v,a,b;
  scanf("%d%d%d%d",&u,&v,&a,&b);
  es[u].emplace_back(v,a);
  es[v].emplace_back(u,a);
  et[u].emplace_back(v,b);
  et[v].emplace_back(u,b);
 }
 sh(es,s,ds);sh(et,t,dt);
 for(int i=1;i<=n;++i)q.emplace_back(ms.insert(ds[i]+dt[i]));
 for(int i=0;i<n;++i){
  printf("%lld\n",int64_t(1e15)-*ms.begin());
  ms.erase(q[i]);
 }
}