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<cstdio>
#include<vector>
#include<queue>
#include<utility>
using namespace std;

struct edge {int to, cost;};
typedef pair<long long, int> P;

const int MAX_N = 100000;
long long dy[MAX_N];
long long ds[MAX_N];
vector<edge> Gy[MAX_N];
vector<edge> Gs[MAX_N];
long long ans[MAX_N];
int n;
const long long INF = 1000000000000000;

void add_edge(int u, int v, int a, int b) {
	Gy[u].push_back({v, a});
	Gy[v].push_back({u, a});
	Gs[u].push_back({v, b});
	Gs[v].push_back({u, b});
}

void dijkstra(vector<edge> G[], long long d[], int s) {
	priority_queue<P, vector<P>, greater<P> > que;
	fill(d, d+n, INF);
	d[s] = 0;
	que.push(P(0, s));
	while(!que.empty()) {
		P p = que.top(); que.pop();
		int v = p.second;
		if(d[v] < p.first) continue;
		for(int i=0;i<G[v].size();i++) {
			edge e = G[v][i];
			if(d[e.to] > d[v] + e.cost) {
				d[e.to] = d[v] + e.cost;
				que.push(P(d[e.to], e.to));
			}
		}
	}
}

int main() {
	int m, s, t;
	scanf("%d %d %d %d", &n, &m, &s, &t);
	s--; t--;
	for(int i=0;i<m;i++) {
		int u, v, a, b;
		scanf("%d %d %d %d", &u, &v, &a, &b);
		add_edge(u-1, v-1, a, b);
	}
	dijkstra(Gy, dy, s);
	dijkstra(Gs, ds, t);
	for(int i=n-1;i>=0;i--) {
		ans[i] = dy[i] + ds[i];
		if(i < n-1 && ans[i] > ans[i+1]) ans[i] = ans[i+1];
	}
	for(int i=0;i<n;i++) {
		printf("%lld\n", INF - ans[i]);
	}
	return 0;
}
