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<queue>
#include<cstdio>
#include<vector>
#include<cstring>
#include<algorithm>
using namespace std;
#define ll long long
const int MAXN=100005;
const ll INF=1000000000000000;
int n,m,s,t;
queue<int>T;
struct node{
	int go,vis;ll dist[5];
	vector<ll>T[5],P;
}w[MAXN];
ll ans[MAXN];
void Add(int a,int b,int c,int d){
	w[a].T[1].push_back(c);
	w[a].T[2].push_back(d);
	w[a].P.push_back(b);
}
void SPFA(int x,int y){
	while(!T.empty())T.pop();
	for(int i=0;i<=n+1;i++)
		w[i].vis=0,w[i].dist[y]=INF;
	T.push(x);w[x].dist[y]=0;
	while(!T.empty()){
		int u=T.front();T.pop();
		w[u].vis=0;
		for(int i=0;i<w[u].P.size();i++){
			int v=w[u].P[i],e=w[u].T[y][i];
			if(e+w[u].dist[y]<w[v].dist[y]){
				w[v].dist[y]=e+w[u].dist[y];
				if(!w[v].vis){
					w[v].vis=1;
					T.push(v);
				}
			}
		}
	}
}
int main(){
	scanf("%d%d%d%d",&n,&m,&s,&t);
	for(int i=1;i<=m;i++){
		int a,b,c,d;
		scanf("%d%d%d%d",&a,&b,&c,&d);
		Add(a,b,c,d);Add(b,a,c,d);
	}
	SPFA(s,1);SPFA(t,2);
	ans[n+1]=INF;
	for(int i=n;i>=1;i--)
		ans[i]=min(ans[i+1],w[i].dist[1]+w[i].dist[2]);
	for(int i=1;i<=n;i++)
		printf("%lld\n",INF-ans[i]);
}
