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
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<long[]>[] rinsetuList = new ArrayList[n];
		ArrayList<long[]>[] rinsetuList2 = new ArrayList[n];
		for (int i = 0; i < n; i++) {
			rinsetuList[i] = new ArrayList<long[]>();
			rinsetuList2[i] = new ArrayList<long[]>();
		}

		//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<long[]> comp = (x, y) -> x[1] < y[1] ? -1 : 1;
		PriorityQueue<long[]> nodeQue = new PriorityQueue<long[]>(comp);
		PriorityQueue<long[]> nodeQue2 = new PriorityQueue<long[]>(comp);
		nodeQue.add(new long[] { s, 0 });
		nodeQue2.add(new long[] { t, 0 });

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

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

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

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

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

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

	}

	private static long[] MyDijkstra(int startIndex, int nodeN, ArrayList<long[]>[] rinsetuList,
			PriorityQueue<long[]> nodeQue) {
		//スタートノードからの道のり
		long[] d = new long[nodeN];
		for (int i = 0; i < nodeN; i++) {
			d[i] = i == startIndex ? 0 : 1000000000000000001L;
		}
		//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
			long[] nearestNode = nodeQue.poll();

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


//		ArrayList<Integer> path = new ArrayList<Integer>();
//		int goalIndex = 4;
//		int i = goalIndex;
//		while (true) {
//			path.add(i);
//			if (i == startIndex)
//				break;
//			i = prev[i];
//		}
//		Collections.reverse(path);

		return d;
		//
		//		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());
	}
}