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 NMAX 100005

typedef long long int ll;

typedef pair<int, int> iPair;
vector<iPair> adjs[NMAX];
vector<iPair> adjt[NMAX];
int N, M;

void pcc(int src, ll dist[], vector<iPair> adj[]){
	dist[src]=0;
	priority_queue <iPair, vector<iPair>, greater<iPair> > pq;
	pq.push(make_pair(0,src));
	while(!pq.empty()){
		int u = pq.top().second;
		pq.pop();
		for(auto x : adj[u]){
			int v = x.second;
			int w = x.first;
			if(dist[v] > dist[u] + w){
				dist[v] =  dist[u]+w;
				pq.push(make_pair(dist[v],v));
			}
		}
	}
}

int main(){
	
	ll yen = pow(10,15);
	
	int s, t; cin>>N>>M>>s>>t;
	for(int i=0;i<M;i++){
		int u, v, a, b;
		cin>>u>>v>>a>>b;
		adjs[u].push_back(make_pair(a,v));
		adjs[v].push_back(make_pair(a,u));
		adjt[u].push_back(make_pair(b,v));
		adjt[v].push_back(make_pair(b,u));
	}
	
	ll dist_s[NMAX];
	ll dist_t[NMAX];
	for(int i=0;i<NMAX;i++){
		dist_s[i] = LONG_LONG_MAX;
		dist_t[i] = LONG_LONG_MAX;
	}
	
	pcc(s, dist_s, adjs);
	pcc(t, dist_t, adjt);
	
	ll ans[N+1];
	ans[N] = dist_s[N] + dist_t[N];
	for(int i=N-1;i>0;i--){
		ans[i] = min(ans[i+1], dist_s[i]+dist_t[i]);
	}
	
	for(int i=1;i<=N;i++){
		printf("%lld\n",yen-ans[i]);
	}
}
