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<vector>
#include<algorithm>
#include<queue>
using namespace std;
#define MAXN 100010
#define INF 1E15
#define LL long long
typedef pair<LL,int> PLI;
struct node
{
	int v,a,b;
};
vector<node>G[MAXN];
int n,m,s,t,u,v,a,b;
bool vis[MAXN];
LL d1[MAXN],d2[MAXN];
LL ans[MAXN];
void Solve()
{
	priority_queue<PLI,vector<PLI>,greater<PLI> >Q;
	fill(d1+1,d1+n+1,INF);
	fill(d2+1,d2+n+1,INF);
	d1[s]=0,d2[t]=0;
	Q.push(PLI(0,s));
	while(!Q.empty())
	{
		PLI q=Q.top();Q.pop();
		int u=q.second;
		if(d1[u]<q.first) continue;
		for(int i=0;i<G[u].size();i++)
		{
			node e=G[u][i];
			if(d1[e.v]>d1[u]+e.a)
			{
				d1[e.v]=d1[u]+e.a;
				Q.push(PLI(d1[e.v],e.v));
			}
		}
	}
	Q.push(PLI(0,t));
	while(!Q.empty())
	{
		PLI q=Q.top();Q.pop();
		int u=q.second;
		if(d2[u]<q.first) continue;
		for(int i=0;i<G[u].size();i++)
		{
			node e=G[u][i];
			if(d2[e.v]>d2[u]+e.b)
			{
				d2[e.v]=d2[u]+e.b;
				Q.push(PLI(d2[e.v],e.v));
			}
		}
	}
	for(int i=1;i<=n;i++)
		ans[i]=d1[i]+d2[i];
	for(int i=n-1;i>=1;i--)
		ans[i]=min(ans[i],ans[i+1]);
}
int main()
{
	node e;
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=0;i<m;i++)
	{
		scanf("%d%d%d%d",&u,&v,&e.a,&e.b);
		e.v=v;G[u].push_back(e);
		e.v=u;G[v].push_back(e);
	}
	Solve();
	for(int i=1;i<=n;i++)
		printf("%lld\n",1000000000000000-ans[i]);
}