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>
#define maxn 200010
#define INF 0x3f3f3f3f 
using namespace std;
struct Edge{
	int to,next,val;
	Edge(int a=0,int b=0,int c=0){
		to=a,next=b,val=c;
	}
}l[maxn<<1][2];
int head[maxn][2],cnt[2],n,m;
long long dis[maxn][2],res[maxn];
void SPFA(int s,int k){
	queue<int>q;int vis[maxn];
	memset(vis,0,sizeof(vis));
	memset(dis,INF,sizeof(dis));
	q.push(s),dis[s][k]=0;
	while(!q.empty()){
		int hq=q.front();q.pop();vis[hq]=0;
		for (int i=head[hq][k];i;i=l[i][k].next){
			int v=l[i][k].to;
			if (dis[v][k]>dis[hq][k]+l[i][k].val){
				dis[v][k]=dis[hq][k]+l[i][k].val;
				if (!vis[v]) vis[v]=1,q.push(v);
			}
		}
	}
	for (int i=1;i<=n;i++) res[i]+=dis[i][k];
}
void add(int a,int b,int c,int k){
	l[++cnt[k]][k]=Edge(b,head[a][k],c);
	head[a][k]=cnt[k];
}
int main(){
	ios::sync_with_stdio(false);
	int s,t; cin>>n>>m>>s>>t;
	for (int i=1;i<=m;i++){
		int x,y,a,b;
		cin>>x>>y>>a>>b;
		add(x,y,a,0),add(y,x,b,1);
		add(y,x,a,0),add(x,y,b,1);
	}
	SPFA(s,0),SPFA(t,1);
	for (int i=n-1;i>=1;i--) res[i]=min(res[i+1],res[i]);
	for (int i=1;i<=n;i++) cout<<1000000000000000-res[i]<<endl;
}