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;
const int maxn=100100;
const long long INF=1e18;
const long long Fir=1e15;
int S,T;
int n,m;
int beg[maxn],tto[maxn<<1],nex[maxn<<1],w1[maxn<<1],w2[maxn<<1],e;
void putin(int s,int t,int x,int y){
	tto[++e]=t;
	nex[e]=beg[s];
	beg[s]=e;
	w1[e]=x;
	w2[e]=y;
}
long long dis1[maxn],dis2[maxn];
struct node{
	int p;
	long long d;
};
bool operator < (const node &A,const node &B){
	return A.d>B.d;
}
priority_queue<node> q;
bool vis[maxn];
void dij(long long *dis,int *w,int s){
	for(int i=1;i<=n;i++)
		dis[i]=INF,vis[i]=0;
	dis[s]=0;
	q.push((node){s,0});
	node st;
	int u;
	while(!q.empty()){
		st=q.top();
		q.pop();
		u=st.p;
		if(vis[u]) continue;
		vis[u]=1;
		for(int i=beg[u];i;i=nex[i]){
			if(dis[tto[i]]>dis[u]+w[i]){
				dis[tto[i]]=dis[u]+w[i];
				q.push((node){tto[i],dis[tto[i]]});
			}
		}
	}
}
long long ans[maxn];
int main(){
//	freopen("A.in","r",stdin);
	scanf("%d%d",&n,&m);
	scanf("%d%d",&S,&T);
	int s,t,x,y;
	for(int i=1;i<=m;i++){
		scanf("%d%d%d%d",&s,&t,&x,&y);
		putin(s,t,x,y);
		putin(t,s,x,y);
	}
	dij(dis1,w1,S);
	dij(dis2,w2,T);
	if(dis1[T]==INF){
		for(int i=0;i<=n-2;i++)
			printf("-1\n");
		return 0;
	}
	for(int i=1;i<=n;i++)
		ans[i]=dis1[i]+dis2[i];
	for(int i=n;i>=1;i--)
		if(ans[i]<ans[i-1])
			ans[i-1]=ans[i];
	for(int i=1;i<=n;i++){
		if(ans[i]<INF) printf("%lld\n",Fir-ans[i]);
		else
			printf("0\n");
	}
	return 0;
}
