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;
#define INF (1<<30)
#define INFLL (1ll<<60)
typedef pair<ll, int> P;
typedef pair<int, P> E;
#define MOD (1000000007ll)
#define l_ength size
#define EPS (1e-10)

void add_mod(ll &a, ll b){
	a += b;
	a %= MOD;
}

void mul_mod(ll &a, ll b){
	a *= b;
	a %= MOD;
}

vector<P> g[2][123456];
priority_queue< P, vector<P>, greater<P> > pq;
ll dist[2][123456];

int main(void){
	int n,m,s[2],u,v,k,i,dummy;
	ll a,b,x,d,c;
	P tmp;
	fill(dist[0],dist[2],INFLL);
	x = (ll)(1e15);
	cin >> n >> m >> s[0] >> s[1]; --s[0]; --s[1];
	for(i=0; i<m; ++i){
		cin >> u >> v >> a >> b; --u; --v;
		g[0][u].push_back(P(a,v));
		g[0][v].push_back(P(a,u));
		g[1][u].push_back(P(b,v));
		g[1][v].push_back(P(b,u));
	}
	for(k=0; k<2; ++k){
		dist[k][s[k]] = 0ll;
		pq.push(P(dist[k][s[k]],s[k]));
		while(!pq.empty()){
			tmp = pq.top();
			pq.pop();
			d = tmp.first;
			u = tmp.second;
			for(i=(g[k][u].l_ength()-1); i>=0; --i){
				v = g[k][u][i].second;
				c = g[k][u][i].first;
				if(d + c < dist[k][v]){
					dist[k][v] = d + c;
					pq.push(P(dist[k][v],v));
				}
			}
		}
	}
	for(i=0; i<n; ++i){
		pq.push(P(dist[0][i]+dist[1][i],i));
	}
	for(i=0; i<n; ++i){
		tmp = pq.top();
		while(tmp.second < i){
			pq.pop();
			tmp = pq.top();
		}
		cout << (x-tmp.first) << endl;
	}
	return 0;
}
