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 rep(i,a,b) for(int i=a;i<=b;++i)
#define rpd(i,a,b) for(int i=a;i>=b;--i)
#define ll long long
#define pb push_back
using namespace std;
const int N=1e5+5;
const ll INF=1e15;
inline int read(){
    int x=0,f=1;char ch=getchar();
    while(!isdigit(ch)){if(ch=='-')f=-1;ch=getchar();}
    while(isdigit(ch)){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m,S,T,tot;
int head[N],nxt[N<<1],to[N<<1],b[N<<1][3];
ll dis[N][3],ans[N],Min=INF;
bool v[N];
void add(int u,int v,int w0,int w1)
{nxt[++tot]=head[u];head[u]=tot;to[tot]=v;b[tot][0]=w0;b[tot][1]=w1;}
void dfs(int s,int k){
	rep(i,1,n)dis[i][k]=INF,v[N]=0;
	dis[s][k]=0;v[s]=1;queue<int>Q;Q.push(s);
	while(!Q.empty()){
		int k1=Q.front();Q.pop();v[k1]=0;
		for(int i=head[k1];i;i=nxt[i]){
			int p=to[i];
			if(dis[p][k]<=dis[k1][k]+b[i][k])continue;
			dis[p][k]=dis[k1][k]+b[i][k];
			if(v[p])continue;v[p]=1;Q.push(p);
		}
	}
}
int main(){
	n=read();m=read();S=read();T=read();
	rep(i,1,m){
	    int u=read(),v=read(),w0=read(),w1=read();
		add(u,v,w0,w1);add(v,u,w0,w1);
	}
	dfs(S,0);dfs(T,1);
	rpd(i,n,1){
		ll now=dis[i][0]+dis[i][1];
		Min=min(Min,now);
		ans[i]=Min;
	}
	rep(i,1,n){
		ll now=1e15;
		printf("%lld\n",now-ans[i]);
	}
	return 0;
}
