Problem:
Score : 400 points

Problem Statement

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.

Sample Input 1

4 3 2 3
1 4 1 100
1 2 1 10
1 3 20 1

Sample Output 1

999999999999998
999999999999989
999999999999979
999999999999897

After 0 years, it is optimal to do the exchange in City 1.

After 1 years, it is optimal to do the exchange in City 2.

Note that City 1 can still be visited even after the exchange office is closed.
Also note that, if it was allowed to keep 1 yen when do the exchange in City 2 and change the remaining yen into snuuk, we could reach City 3 with 999999999999998 snuuk, but this is NOT allowed.

After 2 years, it is optimal to do the exchange in City 3.

After 3 years, it is optimal to do the exchange in City 4.
Note that the same train can be used multiple times.

Sample Input 2

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

Sample Output 2

999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994
999999574976994

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Wrong Answer)

import java.io.IOException;
import java.io.InputStream;
import java.util.ArrayList;
import java.util.Comparator;
import java.util.NoSuchElementException;
import java.util.PriorityQueue;

public class Main {
	public static void main(String[] args) {
		FastScanner fs = new FastScanner();
		int n = fs.nextInt();
		int m = fs.nextInt();
		int s = fs.nextInt() - 1;
		int t = fs.nextInt() - 1;

		//隣接リスト
		//iとjが距離dで接続しているとき、rinsetuList[i]はint[]型の{j ,d}を要素として持つ
		ArrayList<int[]>[] rinsetuList = new ArrayList[n];
		ArrayList<int[]>[] rinsetuList2 = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			rinsetuList[i] = new ArrayList<int[]>();
			rinsetuList2[i] = new ArrayList<int[]>();
		}

		//ノードの優先度付きキュー
		//(nodeIndex, sからの道のり)をsからの道のりが小さい順に格納
		Comparator<int[]> comp = (x, y) -> x[1] - y[1];
		PriorityQueue<int[]> nodeQue = new PriorityQueue<int[]>(comp);
		PriorityQueue<int[]> nodeQue2 = new PriorityQueue<int[]>(comp);
		nodeQue.add(new int[] { s, 0 });
		nodeQue2.add(new int[] { t, 0 });

		for (int i = 0; i < m; i++) {
			int u = fs.nextInt() - 1;
			int v = fs.nextInt() - 1;
			int a = fs.nextInt();
			int b = fs.nextInt();

			rinsetuList[u].add(new int[] { v, a });
			rinsetuList[v].add(new int[] { u, a });
			rinsetuList2[u].add(new int[] { v, b });
			rinsetuList2[v].add(new int[] { u, b });

			if (u == s) {
				nodeQue.add(new int[] { v, a });
			}
			if (v == s) {
				nodeQue.add(new int[] { u, a });
			}
			if (u == s) {
				nodeQue2.add(new int[] { v, b });
			}
			if (v == s) {
				nodeQue2.add(new int[] { u, b });
			}
		}

		int[] res = MyDijkstra(s, n, rinsetuList, nodeQue);
		int[] res2 = MyDijkstra(t, n, rinsetuList2, nodeQue2);

		int[] output = new int[n];
		int min = Integer.MAX_VALUE;
		for(int i = n - 1;i >= 0; i--) {
			if(res[i] + res2[i] < min) {
				min = res[i] + res2[i];
			}else {
			}
			output[i] = min;
		}

		for (int i = 0; i < n; i++) {
			System.out.println(1000000000000000L - output[i]);
		}

	}

	private static int[] MyDijkstra(int startIndex, int nodeN, ArrayList<int[]>[] rinsetuList,
			PriorityQueue<int[]> nodeQue) {
		//スタートノードからの道のり
		int[] d = new int[nodeN];
		for (int i = 0; i < nodeN; i++) {
			d[i] = i == startIndex ? 0 : 1000000001;
		}
		//直前に通るべきノード
//		int[] prev = new int[nodeN];

		while (!nodeQue.isEmpty()) {
			//最も近いノード
			int[] nearestNode = nodeQue.poll();

			for (int[] node : rinsetuList[nearestNode[0]]) {
				int dist = d[nearestNode[0]] + node[1];
				if (d[node[0]] > dist) {
					d[node[0]] = dist;
//					prev[node[0]] = nearestNode[0];
					nodeQue.add(new int[] { node[0], dist });
				}
			}
		}
		return d;

//		ArrayList<Integer> path = new ArrayList<Integer>();
//		int goalIndex = 0;
//		int i = goalIndex;
//		while(true) {
//			path.add(i);
//			if(i == startIndex) break;
//			i = prev[i];
//		}
//		Collections.reverse(path);
//
//		return new ShortestPath(d, path);

	}
}

//class ShortestPath {
//	int[] dist;
//	ArrayList<Integer> path;
//
//	public ShortestPath(int[] dist, ArrayList<Integer> path) {
//		this.dist = dist;
//		this.path = path;
//	}
//}

class FastScanner {
private final InputStream in = System.in;
private final byte[] buffer = new byte[1024];
private int ptr = 0;
private int buflen = 0;
private boolean hasNextByte() {
  if (ptr < buflen) {
      return true;
  }else{
      ptr = 0;
      try {
          buflen = in.read(buffer);
      } catch (IOException e) {
          e.printStackTrace();
      }
      if (buflen <= 0) {
          return false;
      }
  }
  return true;
}
private int readByte() { if (hasNextByte()) return buffer[ptr++]; else return -1;}
private static boolean isPrintableChar(int c) { return 33 <= c && c <= 126;}
public boolean hasNext() { while(hasNextByte() && !isPrintableChar(buffer[ptr])) ptr++; return hasNextByte();}
public String next() {
  if (!hasNext()) throw new NoSuchElementException();
  StringBuilder sb = new StringBuilder();
  int b = readByte();
  while(isPrintableChar(b)) {
      sb.appendCodePoint(b);
      b = readByte();
  }
  return sb.toString();
}
public long nextLong() {
  if (!hasNext()) throw new NoSuchElementException();
  long n = 0;
  boolean minus = false;
  int b = readByte();
  if (b == '-') {
      minus = true;
      b = readByte();
  }
  if (b < '0' || '9' < b) {
      throw new NumberFormatException();
  }
  while(true){
      if ('0' <= b && b <= '9') {
          n *= 10;
          n += b - '0';
      }else if(b == -1 || !isPrintableChar(b)){
          return minus ? -n : n;
      }else{
          throw new NumberFormatException();
      }
      b = readByte();
  }
}
public int nextInt() {
  long nl = nextLong();
  if (nl < Integer.MIN_VALUE || nl > Integer.MAX_VALUE) throw new NumberFormatException();
  return (int) nl;
}
public double nextDouble() { return Double.parseDouble(next());}
}