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>
#define FOR(i,a,b) for(int i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
	
using namespace std;

typedef long long ll;
typedef pair<int,ll> edge;
typedef pair<ll,int> P;
const ll INF=1e18;

const int MAX_N=1e5,MAX_M=1e5;
int N,M,S,T;
vector<edge> G[MAX_N][2];
ll dist[MAX_N][2];

void dijkstra(int i){
	int s;
	if (i==0){
		s=S;
	}else{
		s=T;
	}
	REP(v,N){
		dist[v][i]=INF;
	}
	dist[s][i]=0;
	priority_queue<P,vector<P> ,greater<P> > pque;
	pque.push(P(0,s));
	while(!pque.empty()){
		P p=pque.top();
		pque.pop();
		int v=p.second;
		if (dist[v][i]<p.first){
			continue;
		}
		REP(j,(int)G[v][i].size()){
			edge e=G[v][i][j];
			if (dist[e.first][i]>dist[v][i]+e.second){
				dist[e.first][i]=dist[v][i]+e.second;
				pque.push(P(dist[e.first][i],e.first));
			}
		}
	}
}

int main(){
	cin>>N>>M>>S>>T;
	S--;
	T--;
	REP(i,M){
		int u,v,a,b;
		cin>>u>>v>>a>>b;
		u--;
		v--;
		G[u][0].push_back(edge(v,a));
		G[v][0].push_back(edge(u,a));
		G[u][1].push_back(edge(v,b));
		G[v][1].push_back(edge(u,b));
	}
	dijkstra(0);
	dijkstra(1);
	vector<ll> vans(N);
	ll ans=0;
	for(int i=N-1;i>=0;i--){
		ans=max(ans,(ll)1e15-dist[i][0]-dist[i][1]);
		vans[i]=ans;
	}
	REP(i,N){
		cout<<vans[i]<<endl;
	}
	return 0;
}