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>
#define rep(i,n)for(int i=0;i<(n);i++)
#define MOD 1000000007
#define INF 0x3f3f3f3f
#define INFL 0x3f3f3f3f3f3f3f3f
#define EPS (1e-10)
using namespace std;
typedef long long ll;
typedef pair<ll,ll>P;

vector<P>E[200000],G[200000];
ll d[200000],e[200000];
int main(){
	int n,m,s,t;cin>>n>>m>>s>>t;s--;t--;
	rep(i,m){
		int u,v,a,b;cin>>u>>v>>a>>b;u--;v--;
		E[u].push_back(P(a,v));E[v].push_back(P(a,u));
		G[u].push_back(P(b,v));G[v].push_back(P(b,u));
	}
	priority_queue<P,vector<P>,greater<P>>que;
	memset(d,0x3f,sizeof(d));
	d[s]=0;que.push(P(0,s));
	while(!que.empty()){
		P p=que.top();que.pop();
		if(d[p.second]!=p.first)continue;
		for(auto u:E[p.second]){
			if(d[u.second]>p.first+u.first){
				d[u.second]=p.first+u.first;
				que.push(P(d[u.second],u.second));
			}
		}
	}
	memset(e,0x3f,sizeof(e));
	e[t]=0;que.push(P(0,t));
	while(!que.empty()){
		P p=que.top();que.pop();
		if(e[p.second]!=p.first)continue;
		for(auto u:G[p.second]){
			if(e[u.second]>p.first+u.first){
				e[u.second]=p.first+u.first;
				que.push(P(e[u.second],u.second));
			}
		}
	}
	deque<P>v;
	rep(i,n){
		v.push_back(P(d[i]+e[i],i));
	}
	sort(v.begin(),v.end());
	ll C=1e15;
	rep(i,n){
		while(v.front().second<i)v.pop_front();
		printf("%lld\n",C-v.front().first);
	}
}