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 <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#include <functional>
#include <utility>

using namespace std;
typedef long long lli;
typedef pair<lli,int> pli;

const lli inf=1e15;

struct edge{
	int t,w[2];
	edge() {}
	edge(int t,int a,int b):t(t){
		w[0]=a; w[1]=b;
	}
};

vector<edge> adj[100001];
int n,m,s,t;
lli dist[100001][2],mx[100001];
int vis[100001];

void dijk(int s,int t,int j) {
	priority_queue<pli,vector<pli>,greater<pli>> que;
	que.push(pli(0,s));
	fill(vis,vis+n+1,0);
	for(int i=1;i<=n;i++) dist[i][j] = inf;
	dist[s][j] = 0;

	while(!que.empty()) {
		lli cost=que.top().first;
		int cur=que.top().second;
		que.pop();

		if(vis[cur]) continue;
		vis[cur]=1;

		for(auto &it:adj[cur]) {
			if(!vis[it.t] && dist[it.t][j] > cost + it.w[j]) {
				dist[it.t][j] = cost+it.w[j];
				que.push(pli(dist[it.t][j],it.t));
			}
		}
	}
}


int main() {
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=0,u,v,a,b;i<m;i++) {
		scanf("%d%d%d%d",&u,&v,&a,&b);
		adj[u].push_back(edge(v,a,b));
		adj[v].push_back(edge(u,a,b));
	}

	dijk(s,t,0);
	dijk(t,s,1);

	for(int i=n-1;i>=0;i--) {
		mx[i] = dist[i+1][0] + dist[i+1][1];
		if(i!=n-1) mx[i] = min(mx[i],mx[i+1]);
	}
	for(int i=0;i<n;i++) printf("%lld\n",inf-mx[i]);
	
	return 0;
}
