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;
using ll=long long;
#define rep(i,a,n) for(ll i=a;i<n;i++)
#define ALL(s) s.begin(),s.end()
#define P pair<ll,ll>
#define vl vector<ll>
#define vi vector<int>
#define vvl vector<vector<ll>>
#define vvi vector<vector<int>>
#define print(n) cout<<n<<endl
const int M=100111;
const int inf=1000000007;
const long long INF=1e18;
int dx[4]={-1,0,1,0},dy[4]={0,-1,0,1};
int ddx[8]={1,-1,0,0,1,-1,-1,1},ddy[8]={0,0,1,-1,1,-1,1,-1};

ll n,m,s,t,x,y,a,b;
vector<P> to[2][M];
ll dist[2][M];
priority_queue<P,vector<P>,greater<P>> que[2];
void dijkstra(int z,int start){
	rep(i,0,M)dist[z][i]=INF;
	dist[z][start]=0;
	que[z].push(P(0,start));
	while(que[z].size()){
		P p=que[z].top();que[z].pop();
		ll cost=p.first,now=p.second;
		if(dist[z][now]<cost)continue;
		for(auto e:to[z][now]){
			ll cost=e.first,next=e.second;
			if(dist[z][next]>dist[z][now]+cost){
				dist[z][next]=dist[z][now]+cost;
				que[z].push(P(dist[z][next],next));
			}
		}
	}
}


int main(){
	cin>>n>>m>>s>>t;
	rep(i,0,m){
		cin>>x>>y>>a>>b;
		to[0][x].push_back(P(a,y));
		to[0][y].push_back(P(a,x));
		to[1][x].push_back(P(b,y));
		to[1][y].push_back(P(b,x));
	}
	//to1は円の時のdijkstra.
	dijkstra(0,s);
	//to2の時はスヌークの時のdijkstra.
	dijkstra(1,t);
	ll cur[M];
	fill(cur,cur+M,INF);
	for(int i=n;i>0;i--){
		ll tmp=dist[0][i]+dist[1][i];
		cur[i]=min(cur[i+1],tmp);
	}
	rep(i,1,n+1)print(1000000000000000-cur[i]);
}