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<queue>
#include<vector>
#include<algorithm>
#include<iostream>
using namespace std;
typedef long long ll;
struct edge
{
	ll x,w,c;
	bool operator<(const edge&y)const
	{
		return w>y.w;
	}
	bool operator>(const edge&y)const
	{
		return w<y.w;
	}
}h,v;
ll ans[100005],n,m,ss,t,i,a[100005],x,y,w,c,b[100005],vis1[100005],vis2[100005];
vector<edge> s[100005];
priority_queue<edge> q;
ll minn(ll a,ll b){return a<b?a:b;}
int main()
{
	cin>>n>>m>>ss>>t;
	for(i=1;i<=n;i++)a[i]=b[i]=1e16;
	for(i=1;i<=m;i++)
	{
		cin>>x>>y>>w>>c;
		s[x].push_back((edge){y,w,c});
		s[y].push_back((edge){x,w,c});
	}
	a[ss]=0;
	q.push((edge){ss,0,0});
	while(!q.empty())
	{
		h=q.top();
		q.pop();
		if(vis1[h.x])continue;
		vis1[h.x]=1;
		for(i=0;i<s[h.x].size();i++)
		{
			v=s[h.x][i];
			if(vis1[v.x])continue;
			if(a[v.x]>a[h.x]+v.w)
			{
				a[v.x]=a[h.x]+v.w;
				q.push((edge){v.x,a[v.x],0});
			}
		}
	}
	b[t]=0;
	q.push((edge){t,0,0});
	while(!q.empty())
	{
		h=q.top();
		q.pop();
		if(vis2[h.x])continue;
		vis2[h.x]=1;
		for(i=0;i<s[h.x].size();i++)
		{
			v=s[h.x][i];
			if(vis2[v.x])continue;
			if(b[v.x]>b[h.x]+v.c)
			{
				b[v.x]=b[h.x]+v.c;
				q.push((edge){v.x,b[v.x],0});
			}
		}
	}
	for(i=1;i<=n;i++)a[i]+=b[i];
	ans[n+1]=1e16;
	for(i=n;i>0;i--)
	{
		ans[i]=minn(ans[i+1],a[i]);
	}
	for(i=1;i<=n;i++)cout<<1000000000000000-ans[i]<<endl;
	return 0;
}