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 <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <utility>
#include <cstdio>
#include <vector>
#include <string>
#include <queue>
#include <stack>
#include <cmath>
#include <set>
#include <map>
#define my_abs(x) ((x) < 0 ? -(x) : (x))
#define pb push_back
typedef long long ll; 
std::vector<std::pair<int, int> > adj[100005], rev[100005]; 
ll X[100005], Y[100005], ans[100005];
int n, m, s, t; 
template <typename T>
inline void dij(T &g, int st, ll *dis)
{
	std::priority_queue<std::pair<ll, int> > que; 
	memset(dis, 0x3f, n << 3); 
	dis[st] = 0; 
	que.push({0, st}); 
	while (!que.empty())
	{
		int u = que.top().second; 
		ll d = -que.top().first; 
		que.pop(); 
		if (dis[u] != d)
			continue; 
		for (auto &e : g[u])
		{
			if (dis[e.first] > dis[u] + e.second)
			{
				dis[e.first] = dis[u] + e.second; 
				que.push({-dis[e.first], e.first}); 
			}
		}
	}
}
int main()
{
	// freopen("D.in", "r", stdin); 
	scanf("%d%d%d%d", &n, &m, &s, &t);
	for (int i = 0; i < m; i++)
	{
		int u, v, a, b; 
		scanf("%d%d%d%d", &u, &v, &a, &b); 
		adj[--u].pb({--v, a}); 
		adj[v].pb({u, a}); 
		rev[u].pb({v, b}); 
		rev[v].pb({u, b}); 
	}
	dij(adj, --s, X); 
	dij(rev, --t, Y);
	for (int i = n - 1; i >= 0; i--)
		ans[i] = std::max(ans[i + 1], (ll)1e15 - X[i] - Y[i]); 
	for (int i = 0; i < n; i++)
		printf("%lld\n", ans[i]);
	return 0; 
}
