#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> ii;
typedef long long ll;
typedef pair <ll, int> lli;

const int Maxn = 100005;
const ll has = 1000000000000000ll;

int n, m, s, t;
vector <ii> neighS[Maxn], neighT[Maxn];
ll distS[Maxn], distT[Maxn];
ll mx;
ll res[Maxn];

void Dijkstra(int v, vector <ii> neigh[], ll dist[])
{
	fill(dist, dist + Maxn, has); dist[v] = 0;
	priority_queue <lli> Q; Q.push(lli(-dist[v], v));
	while (!Q.empty()) {
		int v = Q.top().second;
		ll d = -Q.top().first; Q.pop();
		if (dist[v] != d) continue;
		for (int i = 0; i < neigh[v].size(); i++) {
			ii u = neigh[v][i];
			if (d + u.second < dist[u.first]) {
				dist[u.first] = d + u.second;
				Q.push(lli(-dist[u.first], u.first));
			}
		}
	}
}

int main()
{
	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);
		neighS[u].push_back(ii(v, a));
		neighS[v].push_back(ii(u, a));
		neighT[u].push_back(ii(v, b));
		neighT[v].push_back(ii(u, b));
	}
	Dijkstra(s, neighS, distS);
	Dijkstra(t, neighT, distT);
	for (int i = n - 1; i >= 0; i--) {
		mx = max(mx, has - distS[i + 1] - distT[i + 1]);
		res[i] = mx;
	}
	for (int i = 0; i <= n - 1; i++)
		printf("%lld\n", res[i]);
	return 0;
}
### Prompt

Your task is to create a CPP solution to the following problem:

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

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;

typedef pair <int, int> ii;
typedef long long ll;
typedef pair <ll, int> lli;

const int Maxn = 100005;
const ll has = 1000000000000000ll;

int n, m, s, t;
vector <ii> neighS[Maxn], neighT[Maxn];
ll distS[Maxn], distT[Maxn];
ll mx;
ll res[Maxn];

void Dijkstra(int v, vector <ii> neigh[], ll dist[])
{
	fill(dist, dist + Maxn, has); dist[v] = 0;
	priority_queue <lli> Q; Q.push(lli(-dist[v], v));
	while (!Q.empty()) {
		int v = Q.top().second;
		ll d = -Q.top().first; Q.pop();
		if (dist[v] != d) continue;
		for (int i = 0; i < neigh[v].size(); i++) {
			ii u = neigh[v][i];
			if (d + u.second < dist[u.first]) {
				dist[u.first] = d + u.second;
				Q.push(lli(-dist[u.first], u.first));
			}
		}
	}
}

int main()
{
	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);
		neighS[u].push_back(ii(v, a));
		neighS[v].push_back(ii(u, a));
		neighT[u].push_back(ii(v, b));
		neighT[v].push_back(ii(u, b));
	}
	Dijkstra(s, neighS, distS);
	Dijkstra(t, neighT, distT);
	for (int i = n - 1; i >= 0; i--) {
		mx = max(mx, has - distS[i + 1] - distT[i + 1]);
		res[i] = mx;
	}
	for (int i = 0; i <= n - 1; i++)
		printf("%lld\n", res[i]);
	return 0;
}
```