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 <iostream>
#include <vector>
#include <algorithm>
#include <queue>
#include <climits>
#include <math.h>
 
using namespace std;
typedef unsigned long ul;
typedef pair<long, int> P;

void dj(long *dist, int s, int n, vector<P> *g) {
	bool used[100001];
	for(int i=1;i<=n;i++){
		dist[i]=9223372036854775807;
		used[i]=false;
	}
	priority_queue<P, vector<P>, greater<P> > q;
	q.push(P(0, s));
	while(!q.empty()){
		P tmp = q.top();q.pop();
		int i = tmp.second;
		if(used[i])continue;
		used[i]=true;
		dist[i]=tmp.first;
		for(P e:g[i]){
			long d=tmp.first+e.first;
			if(dist[e.second]<=d)continue;
			q.push(P(d,e.second));
		}
	}
}

int main() {
	int n,m,s,t;
	vector<P> yen[100001],snuke[100001];
	cin>>n>>m>>s>>t;
	for(int i=0;i<m;i++){
		int u,v,a,b;
		cin>>u>>v>>a>>b;
		yen[u].push_back(P(a,v));
		yen[v].push_back(P(a,u));
		snuke[u].push_back(P(b,v));
		snuke[v].push_back(P(b,u));
	}

	long dist1[100001];
	dj(dist1, s, n, yen);

	long dist2[100001];
	dj(dist2, t, n, snuke);

	for(int i=1;i<=n;i++)dist1[i]+=dist2[i];
	for(int i=n-1;i>=0;i--)dist1[i]=min(dist1[i],dist1[i+1]);
	for(int i=1;i<=n;i++){
		cout<<1000000000000000-dist1[i]<<endl;
	}
}