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=1e5+8;
int n,m,s,t;
long long ans[N];
struct node{
    int cnt,head[N];
    long long cost[N];
    bool flag[N];
    struct Edge {int next, to; long long cost;} edge[N*2];
	queue<int> q;
	void add(int from,int to,long long cost)
	{
		edge[++cnt].next=head[from];
		edge[cnt].to=to;
		edge[cnt].cost=cost;
		head[from]=cnt;
	}
	void spfa(int s)
	{
		for(int i=1;i<=n;i++) cost[i]=1000000000000000000;
		memset(flag,0,sizeof(flag));
		q.push(s); flag[s]=1; cost[s]=0;
		while(!q.empty())
		{
			int x=q.front();
			q.pop();
			flag[x]=0;
			for(int i=head[x];i;i=edge[i].next)
			{
				int to=edge[i].to;
				if(cost[to]>cost[x]+edge[i].cost)
				{
					cost[to]=cost[x]+edge[i].cost;
					if(!flag[to]) q.push(to),flag[to]=1;
				}
			}
		}
	}
}s_to_i,i_to_t;
long long Min(long long x,long long y)
{
    if(x>y) return y;
    return x;
}
int main()
{
    cin>>n>>m>>s>>t;
    for(int i=1;i<=m;i++)
    {
        int u,v;
        long long yen,snu;
        cin>>u>>v>>yen>>snu;
        s_to_i.add(u,v,yen);
        i_to_t.add(u,v,snu);
        s_to_i.add(v,u,yen);
        i_to_t.add(v,u,snu);
    }
    s_to_i.spfa(s);
    i_to_t.spfa(t);
    ans[n]=s_to_i.cost[n]+i_to_t.cost[n];
    for(int i=n-1;i>0;i--) ans[i]=Min(ans[i+1],s_to_i.cost[i]+i_to_t.cost[i]);
    for(int i=1;i<=n;i++) printf("%lld\n",1000000000000000-ans[i]);
}
