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;
typedef pair <int, int> pii;
typedef pair <ll, int> city;

const int MAXN = 1e5 + 5;
const ll INF = 1e15;

int N, M, S, T;
vector <pii> E1[MAXN], E2[MAXN];
ll distS[MAXN], distT[MAXN];
ll ans[MAXN];

void load() {
	scanf("%d%d%d%d", &N, &M, &S, &T);
	while (M--) {
		int u, v, a, b;
		scanf("%d%d%d%d", &u, &v, &a, &b);
		E1[u].push_back({v, a});
		E1[v].push_back({u, a});
		E2[u].push_back({v, b});
		E2[v].push_back({u, b});
	}
}

void dijkstra(int src, ll dist[MAXN], vector <pii> E[MAXN]) {
	for (int i = 1; i <= N; i++)
		dist[i] = INF;
	dist[src] = 0;
	
	set <city> curr;
	for (int i = 1; i <= N; i++)
		curr.insert({dist[i], i});
	
	while (!curr.empty()) {
		int x = curr.begin() -> second;
		curr.erase(curr.begin());
		for (auto it : E[x]) {
			ll tmp = dist[x] + it.second;
			if (tmp < dist[it.first]) {
				curr.erase({dist[it.first], it.first});
				dist[it.first] = tmp;
				curr.insert({dist[it.first], it.first});
			}
		}
	}
}

void solve() {
	dijkstra(S, distS, E1);
	dijkstra(T, distT, E2);
		
	for (int i = N; i; i--) {
		ans[i] = distS[i] + distT[i];
		if (i < N)
			ans[i] = min(ans[i], ans[i + 1]);
	}
	
	for (int i = 1; i <= N; i++)
		printf("%lld\n", INF - ans[i]);
}

int main() {
	load();
	solve();
	return 0;
}