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 rep(i,n) for(int i=0;i<n;i++)
#define int long long
typedef pair<int,int> P;
#define INF 1e18
int n,m,start,goal;
int u,v,a,b;
struct edge{int to,cost;};
vector<edge>yen[114514];
vector<edge>snuuk[114514];
int d1[114514];
int d2[114514];
void dijkstra(int s,string S){
	priority_queue<P,vector<P>,greater<P>>q;
	if(S=="yen"){
		fill(d1,d1+n,INF);
		d1[s]=0;
	}
	else{
		fill(d2,d2+n,INF);
		d2[s]=0;
	}
	q.push({s,0});
	while(!q.empty()){
		P p=q.top();q.pop();
		if(S=="yen"){
			if(d1[p.first]<p.second)continue;
			for(int i=0;i<yen[p.first].size();i++){
				edge e=yen[p.first][i];
				if(d1[e.to]>p.second+e.cost){
					d1[e.to]=p.second+e.cost;
					q.push({e.to,d1[e.to]});
				}
			}
		}
		else{
			if(d2[p.first]<p.second)continue;
			for(int i=0;i<snuuk[p.first].size();i++){
				edge e=snuuk[p.first][i];
				if(d2[e.to]>p.second+e.cost){
					d2[e.to]=p.second+e.cost;
					q.push({e.to,d2[e.to]});
				}
			}
		}
	}
}
signed main(){
	cin>>n>>m>>start>>goal;
	rep(i,m){
		cin>>u>>v>>a>>b;
		yen[u-1].push_back({v-1,a});
		yen[v-1].push_back({u-1,a});
		snuuk[u-1].push_back({v-1,b});
		snuuk[v-1].push_back({u-1,b});
	}
	
	dijkstra(start-1,"yen");
	dijkstra(goal-1,"snuuk");
	priority_queue<P,vector<P>,greater<P>>q;
	rep(i,n)q.push({d1[i]+d2[i],i+1});
	rep(i,n){
		P p;
		while(1){
			p=q.top();
			if(p.second>=i+1)break;
			q.pop();
		}
		cout<<(int)pow(10,15)-p.first<<endl;
	}
	return 0;
}
