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;

const int N=100005;
const long long inf=1000000000000000ll;
int n,m,s,t;
long long ans[N];

struct G
{
	int inQ[N],Q[N],st,ed;
	long long dis[N];
	int cnt,h[N],to[N+N],las[N+N],c[N+N];
	void insert(int u,int v,int w)
	{
		las[++cnt]=h[u],h[u]=cnt,to[cnt]=v,c[cnt]=w;
		las[++cnt]=h[v],h[v]=cnt,to[cnt]=u,c[cnt]=w;
	}
	void spfa()
	{
		memset(dis,-1,sizeof(dis));
		int x,i;
		dis[Q[0]]=0,inQ[Q[0]]=1;
		while(st!=ed)
		{
			x=Q[st++],st%=N,inQ[x]=0;
			for(i=h[x];i;i=las[i])
			    if(dis[to[i]]==-1||dis[x]+c[i]<dis[to[i]])
			    {
			    	dis[to[i]]=dis[x]+c[i];
			    	if(!inQ[to[i]])
			    	{
			    		inQ[to[i]]=1;
			    		Q[ed++]=to[i],ed%=N;
			    	}
			    }
		}
	}
}G1,G2;

int main()
{
	scanf("%d%d%d%d",&n,&m,&s,&t);
	int i,u,v,w,x;
	for(i=1;i<=m;i++)
	{
		scanf("%d%d%d%d",&u,&v,&w,&x);
		G1.insert(u,v,w);
		G2.insert(v,u,x);
	}
	G1.st=0,G1.ed=1,G1.Q[0]=s;
	G2.st=0,G2.ed=1,G2.Q[0]=t;
	G1.spfa();
	G2.spfa();
	for(i=1;i<=n;i++)
		ans[i]=G1.dis[i]+G2.dis[i];
	for(i=n-1;i>=1;i--)
	    ans[i]=min(ans[i],ans[i+1]);
	for(i=1;i<=n;i++)
	    printf("%lld\n",inf-ans[i]);
	return 0;
}
