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 MOD 1000000007
#define INF 0x3f3f3f3f
#define EPS 1e-10
#define setNil(x) memset((x),0,sizeof(x))
#define setINF(x) memset((x),63,sizeof(x))
#define setNegINF(x) memset((x),192,sizeof(x)) //NegINF=-INF-1
#define setNeg1(x) memset((x),255,sizeof(x))
#define PB push_back
using namespace std;
typedef long long ll;
typedef pair<ll,int> pii;
const int N=100005;
int n,m,s,t;
ll dy[N],dz[N];
vector<int> g[N],y[N],z[N];
ll ans[N];
priority_queue<pii,vector<pii>,greater<pii> > q;
int main(){
	ios::sync_with_stdio(false);
	cin.tie(0);
	cin>>n>>m>>s>>t;
	for(int i=0;i<m;i++){
		int u,v,cy,cz;
		cin>>u>>v>>cy>>cz;
		g[u].PB(v);g[v].PB(u);
		y[u].PB(cy);y[v].PB(cy);
		z[u].PB(cz);z[v].PB(cz);
	}
	fill(dy+1,dy+n+1,1e15);
	dy[s]=0;
	q.push(pii(0,s));
	while(!q.empty()){
		int u=q.top().second,cost=q.top().first;
		q.pop();
		if(dy[u]<cost)continue;
		for(int i=0;i<g[u].size();i++){
			int v=g[u][i],c=y[u][i];
			if(dy[u]+c<dy[v]){
				dy[v]=dy[u]+c;
				q.push(pii(dy[v],v));
			}
		}
	}
	fill(dz+1,dz+n+1,1e15);
	dz[t]=0;
	q.push(pii(0,t));
	while(!q.empty()){
		int u=q.top().second,cost=q.top().first;
		q.pop();
		if(dz[u]<cost)continue;
		for(int i=0;i<g[u].size();i++){
			int v=g[u][i],c=z[u][i];
			if(dz[u]+c<dz[v]){
				dz[v]=dz[u]+c;
				q.push(pii(dz[v],v));
			}
		}
	}
	for(int i=n;i;i--)ans[i]=max(ans[i+1],ll(1e15)-dy[i]-dz[i]);
	for(int i=1;i<=n;i++)cout<<ans[i]<<endl;
	return 0;
}