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.
#!/usr/bin/env python3
import sys
# import getpass  # not available on codechef
from collections import defaultdict
input = sys.stdin.readline  # to read input quickly

# available on Google, AtCoder Python3, not available on Codeforces
# import numpy as np
# import scipy

M9 = 10**9 + 7  # 998244353
yes, no = "YES", "NO"
# d4 = [(1,0),(0,1),(-1,0),(0,-1)]
# d8 = [(1,0),(1,1),(0,1),(-1,1),(-1,0),(-1,-1),(0,-1),(1,-1)]
# d6 = [(2,0),(1,1),(-1,1),(-2,0),(-1,-1),(1,-1)]  # hexagonal layout
MAXINT = sys.maxsize

# if testing locally, print to terminal with a different color
# OFFLINE_TEST = getpass.getuser() == "hkmac"
OFFLINE_TEST = False  # codechef does not allow getpass
def log(*args):
    if OFFLINE_TEST:
        print('\033[36m', *args, '\033[0m', file=sys.stderr)

def solve(*args):
    # screen input
    if OFFLINE_TEST:
        log("----- solving ------")
        log(*args)
        log("----- ------- ------")
    return solve_(*args)

def read_matrix(rows):
    return [list(map(int,input().split())) for _ in range(rows)]

def read_strings(rows):
    return [input().strip() for _ in range(rows)]

def minus_one(arr):
    return [x-1 for x in arr]

def minus_one_matrix(mrr):
    return [[x-1 for x in row] for row in mrr]


BUFSIZE = 8192

import os
from io import BytesIO, IOBase
class FastIO(IOBase):
    newlines = 0

    def __init__(self, file):
        self._fd = file.fileno()
        self.buffer = BytesIO()
        self.writable = "x" in file.mode or "r" not in file.mode
        self.write = self.buffer.write if self.writable else None

    def read(self):
        while True:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            if not b:
                break
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines = 0
        return self.buffer.read()

    def readline(self):
        while self.newlines == 0:
            b = os.read(self._fd, max(os.fstat(self._fd).st_size, BUFSIZE))
            self.newlines = b.count(b"\n") + (not b)
            ptr = self.buffer.tell()
            self.buffer.seek(0, 2), self.buffer.write(b), self.buffer.seek(ptr)
        self.newlines -= 1
        return self.buffer.readline()

    def flush(self):
        if self.writable:
            os.write(self._fd, self.buffer.getvalue())
            self.buffer.truncate(0), self.buffer.seek(0)


class IOWrapper(IOBase):
    def __init__(self, file):
        self.buffer = FastIO(file)
        self.flush = self.buffer.flush
        self.writable = self.buffer.writable
        self.write = lambda s: self.buffer.write(s.encode("ascii"))
        self.read = lambda: self.buffer.read().decode("ascii")
        self.readline = lambda: self.buffer.readline().decode("ascii")


sys.stdin, sys.stdout = IOWrapper(sys.stdin), IOWrapper(sys.stdout)
input = lambda: sys.stdin.readline().rstrip("\r\n")


# ---------------------------- template ends here ----------------------------

# https://codeforces.com/blog/entry/80158?locale=en
from types import GeneratorType
def bootstrap(f, stack=[]):
    # usage - please remember to YIELD to call and return
    '''
    @bootstrap
    def recurse(n):
        if n <= 0:
            yield 0
        yield (yield recurse(n-1)) + 2

    res = recurse(10**5)
    '''
    def wrappedfunc(*args, **kwargs):
        if stack:
            return f(*args, **kwargs)
        else:
            to = f(*args, **kwargs)
            while True:
                if type(to) is GeneratorType:
                    stack.append(to)
                    to = next(to)
                else:
                    if stack:
                        stack.pop()
                    if not stack:
                        break
                    to = stack[-1].send(to)
            return to
    return wrappedfunc



def solve_():
    # your solution here
    k = int(input())
    mrr = read_matrix(k-1)  # and return as a list of list of int

    # log(k)
    # log(mrr)

    g = defaultdict(set)
    for a,b in mrr:
        g[a].add(b)
        g[b].add(a)

    # del mrr

    visited = set()

    # + number of children that is not a bud
    # - number of buds
    # + 1


    @bootstrap
    def is_bud(cur):
        total_count_buds = 0
        total_children_not_a_bud = 0

        num_children = 0
        num_children_is_bud = 0

        # children_is_bud = []
        for nex in g[cur]:
            if nex in visited:
                continue
            visited.add(nex)
            child_is_bud, count_buds, children_not_a_bud = yield is_bud(nex)

            num_children += 1
            total_count_buds += count_buds
            total_children_not_a_bud += children_not_a_bud

            if child_is_bud:
                num_children_is_bud += 1

        # this is not a bud
        if num_children == 0 or num_children == num_children_is_bud:
            yield False, total_count_buds, total_children_not_a_bud + (num_children - num_children_is_bud)

        # this is a bud
        yield True, total_count_buds + 1, total_children_not_a_bud + (num_children - num_children_is_bud)

    visited.add(1)
    _, total_count_buds, total_children_not_a_bud = is_bud(1)

    # log(num_children_not_a_bud)
    # log(buds)
    # log(non_buds)

    return total_children_not_a_bud - total_count_buds + 1


# for case_num in [0]:  # no loop over test case
# for case_num in range(100):  # if the number of test cases is specified
for case_num in range(int(input())):

    # read line as an integer

    # read line as a string
    # srr = input().strip()

    # read one line and parse each word as a string
    # lst = input().split()

    # read one line and parse each word as an integer
    # a,b,c = list(map(int,input().split()))
    # lst = list(map(int,input().split()))
    # lst = minus_one(lst)

    # read multiple rows
    # arr = read_strings(k)  # and return as a list of str
    # mrr = minus_one_matrix(mrr)

    res = solve()  # include input here

    # print length if applicable
    # print(len(res))

    # parse result
    # res = " ".join(str(x) for x in res)
    # res = "\n".join(str(x) for x in res)
    # res = "\n".join(" ".join(str(x) for x in row) for row in res)

    # print result
    # print("Case #{}: {}".format(case_num+1, res))   # Google and Facebook - case number required

    print(res)
