A tree is a connected graph without cycles. A rooted tree has a special vertex called the root. The parent of a vertex v (different from root) is the previous to v vertex on the shortest path from the root to the vertex v. Children of the vertex v are all vertices for which v is the parent.

A vertex is a leaf if it has no children. We call a vertex a bud, if the following three conditions are satisfied: 

  * it is not a root, 
  * it has at least one child, and 
  * all its children are leaves. 



You are given a rooted tree with n vertices. The vertex 1 is the root. In one operation you can choose any bud with all its children (they are leaves) and re-hang them to any other vertex of the tree. By doing that you delete the edge connecting the bud and its parent and add an edge between the bud and the chosen vertex of the tree. The chosen vertex cannot be the bud itself or any of its children. All children of the bud stay connected to the bud.

What is the minimum number of leaves it is possible to get if you can make any number of the above-mentioned operations (possibly zero)?

Input

The input consists of multiple test cases. The first line contains a single integer t (1 ≤ t ≤ 10^4) — the number of test cases. Description of the test cases follows.

The first line of each test case contains a single integer n (2 ≤ n ≤ 2 ⋅ 10^5) — the number of the vertices in the given tree.

Each of the next n-1 lines contains two integers u and v (1 ≤ u, v ≤ n, u ≠ v) meaning that there is an edge between vertices u and v in the tree.

It is guaranteed that the given graph is a tree.

It is guaranteed that the sum of n over all test cases doesn't exceed 2 ⋅ 10^5.

Output

For each test case print a single integer — the minimal number of leaves that is possible to get after some operations.

Example

Input


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


Output


2
2
1
2
1

Note

In the first test case the tree looks as follows:

<image>

Firstly you can choose a bud vertex 4 and re-hang it to vertex 3. After that you can choose a bud vertex 2 and re-hang it to vertex 7. As a result, you will have the following tree with 2 leaves:

<image>

It can be proved that it is the minimal number of leaves possible to get.

In the second test case the tree looks as follows:

<image>

You can choose a bud vertex 3 and re-hang it to vertex 5. As a result, you will have the following tree with 2 leaves:

<image>

It can be proved that it is the minimal number of leaves possible to get.
public class E {

	public Object solve () {
		int N = sc.nextInt();
		int [][] E = dec(sc.nextInts(N-1));

		G = graph(N, E);
		B = new boolean [N];
		L = new boolean [N];

		dfs(0, -1);

		int res = 0, R = 0;
		for (int i : rep(N))
			if (L[i]) {
				++R;
				++res;
			}

		for (int i : rep(N))
			if (B[i])
				res += G[i].length - 1;

		for (int i : rep(N)) {
			if (B[i] && R > 0) {
				--R;
				--res;
			}
			R += G[i].length - 1;
		}

		return res;
	}

	void dfs (int s, int p) {
		for (int j : G[s])
			if (j != p)
				dfs(j, s);

		int ch = 0;
		for (int j : G[s])
			if (j != p)
				++ch;

		if (ch == 0)
			L[s] = true;

		if (ch > 0) {
			if (s > 0) {
				boolean BB = true;
				for (int j : G[s])
					if (j != p && !L[j])
						BB = false;
				if (BB) {
					B[s] = true;
					for (int j : G[s])
						if (j != p)
							L[j] = false;
				}
			}

			for (int j : G[s])
				if (j != p)
					if (B[j])
						--ch;
			if (ch == 0)
				L[s] = true;
		}
	}

	boolean [] L, B;

	int [][] G;

	private static final int CONTEST_TYPE = 2;

	private static void init () {
	}

	private static final int INF = (int) 1e9 + 10;
	private static int [][] dec (int [] ... E) { return dec(E, INF); }
	private static int [][] dec (int [][] E, int N) { for (int [] e : E) for (int i = 0; i < e.length && i < N; ++i) --e[i]; return E; }
	private static int [][] dup (int [][] E) {
		int [][] res = new int [2*E.length][];
		for (int i = 0; i < E.length; ++i) {
			res[2*i] = E[i].clone();
			res[2*i+1] = E[i].clone();
			res[2*i+1][0] = E[i][1]; res[2*i+1][1] = E[i][0];
		}
		return res;
	}
	private static int [][][] dwgraph (int N, int [][] E) {
		int [] D = new int [N];
		for (int [] e : E)
			++D[e[0]];
		int [][][] res = new int [2][N][];
		for (int i = 0; i < 2; ++i)
			for (int j = 0; j < N; ++j)
				res[i][j] = new int [D[j]];
		D = new int [N];
		for (int [] e : E) {
			int x = e[0], y = e[1], z = e.length > 2 ? e[2] : 1;
			res[0][x][D[x]] = y;
			res[1][x][D[x]] = z;
			++D[x];
		}
		return res;
	}
	private static int [][] graph (int N, int [][] E) { return wgraph(N, E)[0]; }
	private static int [] rep (int N) { return rep(0, N); }
	private static int [] rep (int S, int T) { if (S >= T) return new int [0]; int [] res = new int [T-S]; for (int i = S; i < T; ++i) res[i-S] = i; return res; }
	private static int [][][] wgraph (int N, int [][] E) { return dwgraph(N, dup(E)); }
	//////////////////////////////////////////////////////////////////////////////////// OFF
	private static IOUtils.MyScanner sc = new IOUtils.MyScanner();
	private static class IOUtils {
		public static class MyScanner {
			public String next () { newLine(); return line[index++]; }
			public int nextInt () { return Integer.parseInt(next()); }
			public String nextLine () { line = null; return readLine(); }
			public String [] nextStrings () { return split(nextLine()); }
			public int [] nextInts () {
				String [] S = nextStrings(); int N = S.length;
				int [] res = new int [N];
				for (int i = 0; i < N; ++i)
					res[i] = Integer.parseInt(S[i]);
				return res;
			}
			public int [][] nextInts (int N) {
				int [][] res = new int [N][];
				for (int i = 0; i < N; ++i)
					res[i] = nextInts();
				return res;
			}
			//////////////////////////////////////////////
			private boolean eol () { return index == line.length; }
			private String readLine () {
				try {
					int b;
					StringBuilder res = new StringBuilder();
					while ((b = System.in.read()) < ' ')
						continue;
					res.append((char)b);
					while ((b = System.in.read()) >= ' ')
						res.append((char)b);
					return res.toString();
				} catch (Exception e) {
					throw new Error (e);
				}
			}
			private MyScanner () {
				try {
					while (System.in.available() == 0)
						Thread.sleep(1);
					start();
				} catch (Exception e) {
					throw new Error(e);
				}
			}
			private String [] line;
			private int index;
			private void newLine () {
				if (line == null || eol()) {
					line = split(readLine());
					index = 0;
				}
			}
			private String [] split (String s) { return s.length() > 0 ? s.split(" ") : new String [0]; }
		}
		private static String build (Object o, Object ... A) { return buildDelim(" ", o, A); }
		private static String buildDelim (String delim, Object o, Object ... A) {
			java.util.List<String> str = new java.util.ArrayList<>();
			append(str, o);
			for (Object p : A)
				append(str, p);
			return String.join(delim, str.toArray(new String [0]));
		}
		//////////////////////////////////////////////////////////////////////////////////
		private static java.text.DecimalFormat formatter = new java.text.DecimalFormat("#.#########");
		private static void start () { if (t == 0) t = millis(); }
		private static void append (java.util.function.Consumer<Object> f, java.util.function.Consumer<Object> g, Object o) {
			if (o.getClass().isArray()) {
				int len = java.lang.reflect.Array.getLength(o);
				for (int i = 0; i < len; ++i)
					f.accept(java.lang.reflect.Array.get(o, i));
			}
			else if (o instanceof Iterable<?>)
				((Iterable<?>)o).forEach(f::accept);
			else
				g.accept(o instanceof Double ? formatter.format(o) : o);
		}
		private static void append (java.util.List<String> str, Object o) {
			append(x -> append(str, x), x -> str.add(x.toString()), o);
		}
		private static java.io.PrintWriter pw = new java.io.PrintWriter(System.out);
		private static void flush () { pw.flush(); System.out.flush(); }
		private static void print (Object o, Object ... A) {
			String res = build(o, A);
			if (DEBUG == 2)
				err(res, '(', time(), ')');
			if (res.length() > 0)
				pw.println(res);
			if (DEBUG == 1)
				flush();
		}
		private static void err (Object o, Object ... A) { System.err.println(build(o, A)); }
		private static int DEBUG;
		private static void exit () {
			String end = "------------------" + System.lineSeparator() + time();
			switch(DEBUG) {
			case 1: print(end); break;
			case 2: err(end); break;
			}
			IOUtils.pw.close();
			System.out.flush();
			System.exit(0);
		}
		private static long t;
		private static long millis () { return System.currentTimeMillis(); }
		private static String time () { return "Time: " + (millis() - t) / 1000.0; }
		private static void run (int N) {
			try { DEBUG = Integer.parseInt(System.getProperties().get("DEBUG").toString()); }
			catch (Throwable t) {}

			for (int n = 1; n <= N; ++n) {
				Object res = new E().solve();
				if (res != null) {
					@SuppressWarnings("all")
					Object o = CONTEST_TYPE == 0 ? "Case #" + n + ": " + build(res) : res;
					print(o);
				}
			}
			exit();
		}
	}
	////////////////////////////////////////////////////////////////////////////////////
	public static void main (String[] args) {
		@SuppressWarnings("all")
		int N = CONTEST_TYPE == 1 ? 1 : sc.nextInt();
		init();
		IOUtils.run(N);
	}
}
