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;
#define X first
#define Y second
typedef long long ll;
typedef pair<ll,ll> P;
const int MAXN=1e5+10;
int n,m,S,T,x,y,a,b;
ll res[MAXN],INF=1e15,d1[MAXN],d2[MAXN];
vector<P> G1[MAXN],G2[MAXN];

void dijkastra(vector<P> *G,ll *d,int start)
{
	priority_queue<P,vector<P>,greater<P> > q;
	for(int i=0;i<MAXN;i++) d[i]=INF;
	q.push(P(0,start));d[start]=0;
	while(!q.empty())
	{
		P t=q.top();q.pop();
		int u=t.Y;ll cost=t.X;
		if(cost>d[u]) continue;
		for(int i=0;i<G[u].size();i++)
		{
			int v=G[u][i].X;
			if(d[v]>d[u]+G[u][i].Y)
				d[v]=d[u]+G[u][i].Y,q.push(P(d[v],v));
		}
	}
}

int main()
{
	scanf("%d%d%d%d",&n,&m,&S,&T);
	for(int i=1;i<=m;i++)
		scanf("%d%d%d%d",&x,&y,&a,&b),
		G1[x].push_back(P(y,a)),G1[y].push_back(P(x,a)),
		G2[x].push_back(P(y,b)),G2[y].push_back(P(x,b));
	
	dijkastra(G1,d1,S);dijkastra(G2,d2,T);
	for(int i=1;i<=n;i++) res[i]=d1[i]+d2[i];
	for(int i=n-1;i>=1;i--) res[i]=min(res[i],res[i+1]);
	for(int i=1;i<=n;i++) printf("%lld\n",INF-res[i]);
	return 0;
}