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<cstdlib>
#include<cstring>
#include<queue>
#include<vector>
#include<algorithm>
#include<set>
using namespace std;
int n,m;
vector<pair<int,int> >G1[100010],G2[100010];
int s,t;
bool vis[100010];
long long ds[100010],dt[100010];
long long alld[100010];
set<pair<long long,int> >st;
void dijkstra(int S,long long dis[],vector<pair<int,int> >G[])
{
	for(int i=1;i<=n;i++)dis[i]=1e18;
	memset(vis,false,sizeof(vis));
	priority_queue<pair<long long,int>, vector<pair<long long,int> > ,greater<pair<long long,int> > >q;
	dis[S]=0LL;
	q.push(make_pair(0,S));
	for(int i=1;i<=n;i++)
	{
		while(!q.empty() && (vis[q.top().second] || dis[q.top().second]!=q.top().first))q.pop();
		int x=q.top().second;
		q.pop();
		vis[x]=true;
		for(int j=0;j<G[x].size();j++)
		{
			int y=G[x][j].first;
			if(vis[y] || dis[x]+G[x][j].second>=dis[y])continue;
			dis[y]=dis[x]+G[x][j].second;
			q.push(make_pair(dis[y],y)); 
		}
	}
}
int main()
{
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;i++)
	{
		int u,v,a,b;
		scanf("%d%d%d%d",&u,&v,&a,&b);
		G1[u].push_back(make_pair(v,a));
		G1[v].push_back(make_pair(u,a));
		G2[u].push_back(make_pair(v,b));
		G2[v].push_back(make_pair(u,b));
	}
	dijkstra(s,ds,G1);
	dijkstra(t,dt,G2);
	for(int i=1;i<=n;i++)alld[i]=ds[i]+dt[i],st.insert(make_pair(alld[i],i));
	for(int i=0;i<n;i++)
	{
		printf("%lld\n",1000000000000000LL-(*st.begin()).first);
		st.erase(st.find(make_pair(alld[i+1],i+1)));
	}
	return 0;
}