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<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define MAXN 100005
#define LL long long
#define INF 1e18
int N,M,S,T,u,v;
LL a,b,f=1e15,ans[MAXN];
LL Min(LL x,LL y)
{
	if(x>y) return y;
		return x;
}
struct graph{
	int cnt,head[MAXN],vis[MAXN];
	LL dis[MAXN];
	struct Edge
	{
		int next,to;LL dis;
	}edge[2*MAXN];
	queue<int> q;
	
	void addedge(int from,int to,LL dis)
	{
		edge[++cnt].next=head[from];
		edge[cnt].to=to;
		edge[cnt].dis=dis;
		head[from]=cnt;
	}
	
	void spfa(int s)
	{
		for(int i=1;i<=N;i++)
			dis[i]=INF;
		memset(vis,0,sizeof(vis));
		q.push(s); vis[s]=1; dis[s]=0;
		while(!q.empty())
		{
			int x=q.front(); q.pop();
			vis[x]=0;
			for(int i=head[x];i;i=edge[i].next)
			{
				int to=edge[i].to;
				if(dis[to]>dis[x]+edge[i].dis)
				{
					dis[to]=dis[x]+edge[i].dis;
					if(!vis[to]) q.push(to),vis[to]=1;
				}
			}
		}
	}
}g1,g2;
int main()
{
	scanf("%d %d %d %d",&N,&M,&S,&T);
	for(int i=1;i<=M;i++)
	{
		scanf("%d %d %lld %lld",&u,&v,&a,&b);
		g1.addedge(u,v,a); g1.addedge(v,u,a);
		g2.addedge(v,u,b); g2.addedge(u,v,b);
	}
	g1.spfa(S); g2.spfa(T);
	ans[N+1]=INF;
	for(int i=N;i>=1;i--) ans[i]=Min(ans[i+1],g1.dis[i]+g2.dis[i]);
	for(int i=1;i<=N;i++)
		printf("%lld\n",f-ans[i]);
	return 0;
}