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 int maxn=1e5+10;
int n,m,s,t;
ll diss[maxn],dist[maxn],ans[maxn];
int ebeg[maxn],eto[maxn<<1],enxt[maxn<<1],ecnt,ew[maxn<<1],eww[maxn<<1];
bool inq[maxn];

void ae(int u,int v,int a,int b){
	++ecnt;
	enxt[ecnt]=ebeg[u];
	ebeg[u]=ecnt;
	eto[ecnt]=v;
	ew[ecnt]=a;
	eww[ecnt]=b;
}

int main(){
	cin>>n>>m>>s>>t;
	for(int i=0,u,v,a,b;i<m;++i){
		cin>>u>>v>>a>>b;
		ae(u,v,a,b);ae(v,u,a,b);
	}
	{
		queue<int> q;
		memset(diss,100,sizeof(diss));
		diss[s]=0;
		q.push(s);
		inq[s]=true;
		while(!q.empty()){
			int pos=q.front();q.pop();
			for(int i=ebeg[pos],v;i;i=enxt[i])
				if(diss[v=eto[i]]>diss[pos]+ew[i]){
					diss[v]=diss[pos]+ew[i];
					if(!inq[v]){
						inq[v]=true;
						q.push(v);
					}
				}
			inq[pos]=false;
		}
		memset(dist,100,sizeof(dist));
		dist[t]=0;
		q.push(t);
		inq[t]=true;
		while(!q.empty()){
			int pos=q.front();q.pop();
			for(int i=ebeg[pos],v;i;i=enxt[i])
				if(dist[v=eto[i]]>dist[pos]+eww[i]){
					dist[v]=dist[pos]+eww[i];
					if(!inq[v]){
						inq[v]=true;
						q.push(v);
					}
				}
			inq[pos]=false;
		}
	}
	for(int i=1;i<=n;++i)
		ans[i]=(ll)1e15-diss[i]-dist[i];
	for(int i=n-1;i;--i)
		ans[i]=max(ans[i],ans[i+1]);
	for(int i=1;i<=n;++i)
		cout<<ans[i]<<endl;
	return 0;
}