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 ll long long
#define pii pair<ll,ll>
#define vii vector<pii>
vii a[100005],b[100005];
ll arr[100005],ans[100005];
//ll a[100005];
//ll b[100005];
bool vis[100005];
ll dis1[100005],dis2[100005];
int main(){
priority_queue<pii,vii,greater<pii>> q;
int n,m,s,t;
cin>>n>>m>>s>>t;
for(int i=1;i<=n;i++){
	vis[i]=0;
	dis1[i]=1e18;
	dis2[i]=1e18;
}
while(m--){
	int u,v;
	ll c,d;
	cin>>u>>v>>c>>d;
	a[u].push_back(make_pair(c,v));
	a[v].push_back(make_pair(c,u));
	b[u].push_back(make_pair(d,v));
	b[v].push_back(make_pair(d,u));
}
//1
dis1[s]=0;
q.push(make_pair(0,s));
while(!q.empty()){
	pii tmp=q.top();
	int u=tmp.second;
	q.pop();
	if(vis[u])
	continue;
	vis[u]=1;
	for(int i=0;i<a[u].size();i++){
		int v=a[u][i].second;
		ll e=a[u][i].first;
		if(dis1[u]+e<dis1[v]){
			dis1[v]=dis1[u]+e;
			q.push(make_pair(dis1[v],v));
		}
	}
}
for(int i=1;i<=n;i++)
vis[i]=0;
dis2[t]=0;
q.push(make_pair(0,t));
while(!q.empty()){
	pii tmp=q.top();
	int u=tmp.second;
	q.pop();
	if(vis[u])
	continue;
	vis[u]=1;
	for(int i=0;i<b[u].size();i++){
		int v=b[u][i].second;
		ll e=b[u][i].first;
		if(dis2[u]+e<dis2[v]){
			dis2[v]=dis2[u]+e;
			q.push(make_pair(dis2[v],v));
		}
	}
}
for(int i=1;i<=n;i++){
	arr[i]=dis1[i]+dis2[i];
}
ans[n]=arr[n];
for(int i=n-1;i>0;i--){
	ans[i]=min(ans[i+1],arr[i]);
}
ll y=1e15;
for(int i=1;i<=n;i++){
	cout<<y-ans[i]<<endl;
}



}