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<iostream>
#include<cstring>
#include<cmath>
#include<map>
#include<stack>
#include<queue>
#include<cstdio>
#include<algorithm>
using namespace std;
#define ll long long
#define REP(i,x,y) for(ll i=x;i<=y;i++)
#define rep(i,n) REP(i,1,n)
#define rep0(i,n) REP(i,0,n-1)
#define repG(i,x) for(ll i=pos[x];~i;i=e[i].next)
const ll N=2e5+7;
const ll INF=1e15;
ll pos1[N],pos2[N],dis1[N],dis2[N],ans[N];
bool vis[N];
ll n,m,s,t,num=0;
struct edge{ll v,next,c;}e[N*4];
void add1(ll x,ll y,ll c){e[num]=(edge){y,pos1[x],c}; pos1[x]=num++;}
void add2(ll x,ll y,ll c){e[num]=(edge){y,pos2[x],c}; pos2[x]=num++;}
void spfa(ll x0,ll *pos,ll *dis){
	memset(vis,1,sizeof(vis));
	rep(i,n)dis[i]=INF;
	dis[x0]=0;
	queue<int>q;
	q.push(x0);
	while(!q.empty()){
		ll u=q.front();
		q.pop();
		vis[u]=1;
		repG(i,u){
			if(dis[e[i].v]>dis[u]+e[i].c){
				dis[e[i].v]=dis[u]+e[i].c;
				if(vis[e[i].v]){
					vis[e[i].v]=0;
					q.push(e[i].v);
				}
			}
		}
	}
}
int main(){
	memset(pos1,-1,sizeof(pos1));
	memset(pos2,-1,sizeof(pos2));
	scanf("%lld%lld%lld%lld",&n,&m,&s,&t);
	rep(i,m){
		ll x,y,a,b; scanf("%lld%lld%lld%lld",&x,&y,&a,&b);
		add1(x,y,a);
		add1(y,x,a);
		add2(x,y,b);
		add2(y,x,b);
	}
	spfa(s,pos1,dis1);
	spfa(t,pos2,dis2);
	ans[n+1]=INF;
	for(ll i=n;i;i--)ans[i]=min(ans[i+1],dis1[i]+dis2[i]);
	rep(i,n)printf("%lld\n",INF-ans[i]);
	return 0;
}
