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<vector>
#include<queue>
#include<set>
#define MAXN 100000
#define INF 1000000000000000LL
#define TOT 1000000000000000LL
using namespace std;
typedef long long LL;
struct edge
{
	LL to,val;
	edge(){};
	edge(LL _to,LL _val):to(_to),val(_val){};
};
vector<edge> G[2][MAXN+5];
multiset<LL> ans;
LL dist[2][MAXN+5],times[MAXN+5];
void SPFA(LL s,LL id)
{
	queue<LL> que;
	for(LL i=0;i<=MAXN;i++)
		dist[id][i]=INF;
	dist[id][s]=0;
	que.push(s);
	while(que.empty()==false)
	{
		LL fro=que.front();
		edge e;
		que.pop();
		for(LL i=0;i<G[id][fro].size();i++)
		{
			edge e=G[id][fro][i];
			if(dist[id][e.to]>dist[id][fro]+e.val)
			{
				dist[id][e.to]=dist[id][fro]+e.val;
				que.push(e.to);
			}
		}
	}
}
int main()
{
	LL n,m,s,t;
	scanf("%lld %lld",&n,&m);
	scanf("%lld %lld",&s,&t);
	LL u,v,a,b;
	for(LL i=1;i<=m;i++)
	{
		scanf("%lld %lld %lld %lld",&u,&v,&a,&b);
		G[0][u].push_back(edge(v,a));
		G[0][v].push_back(edge(u,a));
		G[1][u].push_back(edge(v,b));
		G[1][v].push_back(edge(u,b));
	}
	SPFA(s,0);
	SPFA(t,1);
	for(LL i=1;i<=n;i++)
		ans.insert(dist[0][i]+dist[1][i]);
	LL fro;
	for(LL i=0;i<n;i++)
	{
		fro=*ans.begin();
		printf("%lld\n",TOT-1LL*fro);
		ans.erase(ans.find(dist[0][i+1]+dist[1][i+1]));
	}
}