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 <stdio.h>
#include <iostream>
#include <cstring>
#include <algorithm>
#include <queue>
typedef long long ll;
const ll inf=1000000000000000ll;
const int N=100005;
int n,m,S,T,inq[N];
int head[N],now=0;
ll dis1[N],dis2[N];
ll ans[N];
using namespace std;

struct node
{
	int t,w1,w2,next;
}adj[N<<1];
queue<int> Q;
void add(int s,int t,int a,int b)
{
	adj[++now]=(node){t,a,b,head[s]};
	head[s]=now;
}
void spfa1()
{
	int i,s,t;
	for (i=1;i<=n;i++) dis1[i]=inf;
	Q.push(S),inq[S]=1,dis1[S]=0;
	while (!Q.empty())
	{
		s=Q.front(),Q.pop();inq[s]=0;
		for (i=head[s];i;i=adj[i].next)
		{
			t=adj[i].t;
			if (dis1[t]>dis1[s]+adj[i].w1)
			{
				dis1[t]=dis1[s]+adj[i].w1;
				if (!inq[t]) inq[t]=1,Q.push(t);
			}
		}
	}
}
void spfa2()
{
	int i,s,t;
	for (i=1;i<=n;i++) dis2[i]=inf;
	Q.push(T),inq[T]=1,dis2[T]=0;
	while (!Q.empty())
	{
		s=Q.front(),Q.pop();inq[s]=0;
		for (i=head[s];i;i=adj[i].next)
		{
			t=adj[i].t;
			if (dis2[t]>dis2[s]+adj[i].w2)
			{
				dis2[t]=dis2[s]+adj[i].w2;
				if (!inq[t]) inq[t]=1,Q.push(t);
			}
		}
	}
}
int main()
{
	#ifdef Kay
		freopen ("code.in","r",stdin);
		freopen ("code.out","w",stdout);
	#endif
	scanf ("%d %d %d %d",&n,&m,&S,&T);
	int i,s,t,a,b;
	for (i=1;i<=m;i++)
	{
		scanf ("%d %d %d %d",&s,&t,&a,&b);
		add(s,t,a,b),add(t,s,a,b);
	}
	spfa1(),spfa2();
	for (i=1;i<=n;i++) ans[i]=dis1[i]+dis2[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;
}
