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<cmath>
#include<algorithm>
#include<utility>
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
typedef long long ll;
const int maxn = 1e5 + 233;
vector<int> g[maxn],wa[maxn],wb[maxn];
ll fa[maxn],fb[maxn],res[maxn];
queue<int> q;
int n,m,s,t;
void spfa1()
{
	for(int i=1;i<=n;i++)fa[i] = 1LL<<60;
	fa[s]=0;q.push(s);
	while(!q.empty())
	{
		int x = q.front();q.pop();
		for(int i=0;i<g[x].size();i++)
		{
			int y = g[x][i];
			if(fa[y]>fa[x]+wa[x][i])
			{
				fa[y] = fa[x]+wa[x][i];
				q.push(y);
			}
		}
	}
	return ;
}
void spfa2()
{
	for(int i=1;i<=n;i++)fb[i] = 1LL<<60;
	fb[t]=0;q.push(t);
	while(!q.empty())
	{
		int x = q.front();q.pop();
		for(int i=0;i<g[x].size();i++)
		{
			int y = g[x][i];
			if(fb[y]>fb[x]+wb[x][i])
			{
				fb[y] = fb[x]+wb[x][i];
				q.push(y);
			}
		}
	}
	return ;
}
int main()
{
	cin>>n>>m>>s>>t;
	for(int i=1;i<=m;i++)
	{
		int x,y,a,b;
		scanf("%d%d%d%d",&x,&y,&a,&b);
		g[x].push_back(y);
		g[y].push_back(x);
		wa[x].push_back(a);
		wa[y].push_back(a);
		wb[x].push_back(b);
		wb[y].push_back(b);
	}
	spfa1();spfa2();
	ll ans = 1LL<<60;
	for(int i=n;i;i--)
	{
//		cout<<"!"<<fa[i]<<" "<<fb[i]<<endl;
		ans = min(ans,fa[i]+fb[i]);
//		cout<<"!"<<ans<<endl;
		res[i] = 1e15 - ans;
	}
	for(int i=1;i<=n;i++)printf("%lld\n",res[i]);
	return 0;
}