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;
typedef long long ll;
const ll inf=1000000000000000ll;
#define N 100010
int i,j,k,n,m,x,y,s,t;
int fi[N*2],ne[N*2],la[N*2],a[N*2],c1[N*2],c2[N*2],b[N];
ll d1[N],d2[N];
queue<int>q;
int read(){
	int x=0,f=1;
	char ch=getchar();
	while (ch<'0'||ch>'9'){f=ch=='-'?-f:f;ch=getchar();}
	while (ch>='0'&&ch<='9'){x=x*10+ch-48;ch=getchar();}
	return x*f;
}
void add(int x,int y,int t1,int t2){
	k++;a[k]=y;c1[k]=t1;c2[k]=t2;
	if (fi[x]==0)fi[x]=k;else ne[la[x]]=k;
	la[x]=k;
}
void spfa1(){
	for (i=1;i<=n;i++)d1[i]=inf;d1[s]=0;
	q.push(s);
	while (!q.empty()){
		x=q.front();b[x]=0;q.pop();
		for (int i=fi[x];i;i=ne[i])
			if (d1[a[i]]>d1[x]+c1[i]){
				d1[a[i]]=d1[x]+c1[i];
				if (!b[a[i]]){
					b[a[i]]=0;q.push(a[i]);
				}
			}
	}
}
void spfa2(){
    for (i=1;i<=n;i++)d2[i]=inf;d2[t]=0;
    q.push(t); 
    while (!q.empty()){ 
        x=q.front();b[x]=0;q.pop();
        for (int i=fi[x];i;i=ne[i])
            if (d2[a[i]]>d2[x]+c2[i]){
                d2[a[i]]=d2[x]+c2[i]; 
                if (!b[a[i]]){
                    b[a[i]]=0;q.push(a[i]); 
                } 
            } 
    } 
}
ll ans[N];
int main(){
	n=read();m=read();s=read();t=read();
	while (m--){
		int t1,t2;
		x=read();y=read();t1=read();t2=read();
		add(x,y,t1,t2);add(y,x,t1,t2);
	}
	spfa1();spfa2();
	for (i=n;i>=1;i--)ans[i]=max(ans[i+1],inf-d1[i]-d2[i]);
	for (i=1;i<=n;i++)printf("%lld\n",ans[i]);
	return 0;
}
