#include <bits/stdc++.h>
#define int long long
using namespace std;
const int INF = 1e9 + 7, MOD = 1e9 + 7;
const long long LINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
typedef pair<int, int> P;
int n, m, s, t;
struct edge {
	int to;
	int cost;
};
vector<edge> G1[1000001], G2[1000001];
int D1[1000001], D2[1000001];
void dijkstra(int s, vector<edge>* G, int* d);
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> m >> s >> t;
	int u, v, a, b;
	for (int i = 0; i < m; i++) {
		cin >> u >> v >> a >> b;
		G1[u].push_back({v, a});
		G1[v].push_back({u, a});
		G2[u].push_back({v, b});
		G2[v].push_back({u, b});
	}
	dijkstra(s, G1, D1);
	dijkstra(t, G2, D2);
	vector<int> Ans;
	priority_queue< int, vector<int>, greater<int> > que;
	for (int i = n; i >= 1; i--) {
		que.push(D1[i] + D2[i]);
		Ans.push_back(que.top());
	}
	reverse(Ans.begin(), Ans.end());
	for (auto e : Ans) cout << 1000000000000000 - e << endl;
	return 0;
}

void dijkstra(int s, vector<edge>* G, int* d) {
	priority_queue<P, vector<P>, greater<P>> que;
	fill(d, d + n + 1, LINF);
	d[s] = 0;
	que.push(make_pair(0, s));
	while (que.size()) {
		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));
			}
		}
	}
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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>
#define int long long
using namespace std;
const int INF = 1e9 + 7, MOD = 1e9 + 7;
const long long LINF = 1e18;
const int dx[4] = {1, 0, -1, 0}, dy[4] = {0, 1, 0, -1};
typedef pair<int, int> P;
int n, m, s, t;
struct edge {
	int to;
	int cost;
};
vector<edge> G1[1000001], G2[1000001];
int D1[1000001], D2[1000001];
void dijkstra(int s, vector<edge>* G, int* d);
signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	cin >> n >> m >> s >> t;
	int u, v, a, b;
	for (int i = 0; i < m; i++) {
		cin >> u >> v >> a >> b;
		G1[u].push_back({v, a});
		G1[v].push_back({u, a});
		G2[u].push_back({v, b});
		G2[v].push_back({u, b});
	}
	dijkstra(s, G1, D1);
	dijkstra(t, G2, D2);
	vector<int> Ans;
	priority_queue< int, vector<int>, greater<int> > que;
	for (int i = n; i >= 1; i--) {
		que.push(D1[i] + D2[i]);
		Ans.push_back(que.top());
	}
	reverse(Ans.begin(), Ans.end());
	for (auto e : Ans) cout << 1000000000000000 - e << endl;
	return 0;
}

void dijkstra(int s, vector<edge>* G, int* d) {
	priority_queue<P, vector<P>, greater<P>> que;
	fill(d, d + n + 1, LINF);
	d[s] = 0;
	que.push(make_pair(0, s));
	while (que.size()) {
		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));
			}
		}
	}
}
```