Problem: Arthur owns a ski resort on a mountain. There are n landing spots on the mountain numbered from 1 to n from the top to the foot of the mountain. The spots are connected with one-directional ski tracks. All tracks go towards the foot of the mountain, so there are no directed cycles formed by the tracks. There are at most two tracks leaving each spot, but many tracks may enter the same spot.

A skier can start skiing from one spot and stop in another spot if there is a sequence of tracks that lead from the starting spot and end in the ending spot. Unfortunately, recently there were many accidents, because the structure of the resort allows a skier to go through dangerous paths, by reaching high speed and endangering himself and the other customers. Here, a path is called dangerous, if it consists of at least two tracks.

Arthur wants to secure his customers by closing some of the spots in a way that there are no dangerous paths in the resort. When a spot is closed, all tracks entering and leaving that spot become unusable. 

Formally, after closing some of the spots, there should not be a path that consists of two or more tracks.

Arthur doesn't want to close too many spots. He will be happy to find any way to close at most 4/7n spots so that the remaining part is safe. Help him find any suitable way to do so.

Input

The first line contains a single positive integer T — the number of test cases. T test case description follows.

The first line of each description contains two integers n and m (1 ≤ n ≤ 2 ⋅ 10^5) — the number of landing spots and tracks respectively.

The following m lines describe the tracks. Each of these lines contains two integers x and y (1 ≤ x < y ≤ n) — indices of the starting and finishing spots for the respective track. It is guaranteed that at most two tracks start at each spot. There may be tracks in which starting and finishing spots both coincide.

It is guaranteed that the sum of n over all test cases does not exceed 2 ⋅ 10^5.

Output

For each test case, print a single integer k (0 ≤ k ≤ 4/7n) — the number of spots to be closed. In the next line, print k distinct integers — indices of all spots to be closed, in any order.

If there are several answers, you may output any of them. Note that you don't have to minimize k. It can be shown that a suitable answer always exists.

Example

Input


2
4 6
1 2
1 3
2 3
2 4
3 4
3 4
7 6
1 2
1 3
2 4
2 5
3 6
3 7


Output


2
3 4 
4
4 5 6 7 

Note

In the first sample case, closing any two spots is suitable.

In the second sample case, closing only the spot 1 is also suitable.

Solution (in JAVA):

Solution:
import java.io.ByteArrayInputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.InputMismatchException;
import java.util.List;

public class Main {

	private static final String NO = "NO";
	private static final String YES = "YES";
	InputStream is;
	PrintWriter out;
	String INPUT = "";

	private static final long MOD = 1000000007;
	private static final long inf = 100000000000L;

	void solve() {
		int T = ni();
		for (int i = 0; i < T; i++)
			solve(i);
	}

	void solve(int T) {
		int n = ni();

		int dis[] = new int[n+1], ls[] = new int[n+1], rs[] = new int[n+1];

		int m = ni();
		for (int i = 0; i < m; i++) {
			int u = ni(), v = ni();
			if (ls[u] > 0)
				rs[u] = v;
			else
				ls[u] = v;
		}

		List<Integer> ans = new ArrayList<Integer>();
		for (int i = 1; i <= n; i++) {
			if (dis[i] == 2)
				ans.add(i);
			else {
				dis[ls[i]] = Math.max(dis[ls[i]], dis[i] + 1);
				dis[rs[i]] = Math.max(dis[rs[i]], dis[i] + 1);
			}
//			tr(i, dis);
		}
		out.println(ans.size());
		for (int i : ans)
			out.print(i + " ");
		out.println();
	}

	// a^b
	long power(long a, long b) {
		long x = 1, y = a;
		while (b > 0) {
			if (b % 2 != 0) {
				x = (x * y) % MOD;
			}
			y = (y * y) % MOD;
			b /= 2;
		}
		return x % MOD;
	}

	private long gcd(long a, long b) {
		while (a != 0) {
			long tmp = b % a;
			b = a;
			a = tmp;
		}
		return b;
	}

	void run() throws Exception {
		is = INPUT.isEmpty() ? System.in : new ByteArrayInputStream(INPUT.getBytes());
		out = new PrintWriter(System.out);

		long s = System.currentTimeMillis();
		solve();
		out.flush();
		if (!INPUT.isEmpty())
			tr(System.currentTimeMillis() - s + "ms");
	}

	public static void main(String[] args) throws Exception {
		new Main().run();
	}

	private byte[] inbuf = new byte[1024];
	public int lenbuf = 0, ptrbuf = 0;
	private boolean vis[];

	private int readByte() {
		if (lenbuf == -1)
			throw new InputMismatchException();
		if (ptrbuf >= lenbuf) {
			ptrbuf = 0;
			try {
				lenbuf = is.read(inbuf);
			} catch (IOException e) {
				throw new InputMismatchException();
			}
			if (lenbuf <= 0)
				return -1;
		}
		return inbuf[ptrbuf++];
	}

	private boolean isSpaceChar(int c) {
		return !(c >= 33 && c <= 126);
	}

	private int skip() {
		int b;
		while ((b = readByte()) != -1 && isSpaceChar(b))
			;
		return b;
	}

	private double nd() {
		return Double.parseDouble(ns());
	}

	private char nc() {
		return (char) skip();
	}

	private String ns() {
		int b = skip();
		StringBuilder sb = new StringBuilder();
		while (!(isSpaceChar(b))) { // when nextLine, (isSpaceChar(b) && b != '
									// ')
			sb.appendCodePoint(b);
			b = readByte();
		}
		return sb.toString();
	}

	private char[] ns(int n) {
		char[] buf = new char[n];
		int b = skip(), p = 0;
		while (p < n) {
			if (!(isSpaceChar(b)))
				buf[p++] = (char) b;
			b = readByte();
		}
		return n == p ? buf : Arrays.copyOf(buf, p);
	}

	private char[][] nm(int n, int m) {
		char[][] map = new char[n][];
		for (int i = 0; i < n; i++)
			map[i] = ns(m);
		return map;
	}

	private int[] na(int n) {
		int[] a = new int[n];
		for (int i = 0; i < n; i++)
			a[i] = ni();
		return a;
	}

	private List<Integer> na2(int n) {
		List<Integer> a = new ArrayList<Integer>();
		for (int i = 0; i < n; i++)
			a.add(ni());
		return a;
	}

	private int[][] na(int n, int m) {
		int[][] a = new int[n][];
		for (int i = 0; i < n; i++)
			a[i] = na(m);
		return a;
	}

	private int ni() {
		int num = 0, b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private long[] nl(int n) {
		long[] a = new long[n];
		for (int i = 0; i < n; i++)
			a[i] = nl();
		return a;
	}

	private long[][] nl(int n, int m) {
		long[][] a = new long[n][];
		for (int i = 0; i < n; i++)
			a[i] = nl(m);
		return a;
	}

	private long nl() {
		long num = 0;
		int b;
		boolean minus = false;
		while ((b = readByte()) != -1 && !((b >= '0' && b <= '9') || b == '-'))
			;
		if (b == '-') {
			minus = true;
			b = readByte();
		}

		while (true) {
			if (b >= '0' && b <= '9') {
				num = num * 10 + (b - '0');
			} else {
				return minus ? -num : num;
			}
			b = readByte();
		}
	}

	private static void tr(Object... o) {
		System.out.println(Arrays.deepToString(o));
	}

	public class Pair<K, V> {

		/**
		 * Key of this <code>Pair</code>.
		 */
		private K key;

		/**
		 * Gets the key for this pair.
		 * 
		 * @return key for this pair
		 */
		public K getKey() {
			return key;
		}

		/**
		 * Value of this this <code>Pair</code>.
		 */
		private V value;

		/**
		 * Gets the value for this pair.
		 * 
		 * @return value for this pair
		 */
		public V getValue() {
			return value;
		}

		/**
		 * Creates a new pair
		 * 
		 * @param key   The key for this pair
		 * @param value The value to use for this pair
		 */
		public Pair(K key, V value) {
			this.key = key;
			this.value = value;
		}

		/**
		 * <p>
		 * <code>String</code> representation of this <code>Pair</code>.
		 * </p>
		 *
		 * <p>
		 * The default name/value delimiter '=' is always used.
		 * </p>
		 *
		 * @return <code>String</code> representation of this <code>Pair</code>
		 */
		@Override
		public String toString() {
			return key + "=" + value;
		}

		/**
		 * <p>
		 * Generate a hash code for this <code>Pair</code>.
		 * </p>
		 *
		 * <p>
		 * The hash code is calculated using both the name and the value of the
		 * <code>Pair</code>.
		 * </p>
		 *
		 * @return hash code for this <code>Pair</code>
		 */
		@Override
		public int hashCode() {
			// name's hashCode is multiplied by an arbitrary prime number (13)
			// in order to make sure there is a difference in the hashCode between
			// these two parameters:
			// name: a value: aa
			// name: aa value: a
			return key.hashCode() * 13 + (value == null ? 0 : value.hashCode());
		}

		/**
		 * <p>
		 * Test this <code>Pair</code> for equality with another <code>Object</code>.
		 * </p>
		 *
		 * <p>
		 * If the <code>Object</code> to be tested is not a <code>Pair</code> or is
		 * <code>null</code>, then this method returns <code>false</code>.
		 * </p>
		 *
		 * <p>
		 * Two <code>Pair</code>s are considered equal if and only if both the names and
		 * values are equal.
		 * </p>
		 *
		 * @param o the <code>Object</code> to test for equality with this
		 *          <code>Pair</code>
		 * @return <code>true</code> if the given <code>Object</code> is equal to this
		 *         <code>Pair</code> else <code>false</code>
		 */
		@Override
		public boolean equals(Object o) {
			if (this == o)
				return true;
			if (o instanceof Pair) {
				Pair pair = (Pair) o;
				if (key != null ? !key.equals(pair.key) : pair.key != null)
					return false;
				if (value != null ? !value.equals(pair.value) : pair.value != null)
					return false;
				return true;
			}
			return false;
		}
	}

}