## Problem Statement->

directed tree is a directed graph such that if all edges were undirected, this graph would be a tree. A rooted directed tree is a directed tree in which there is one vertex (the root, let's denote it by $r$) such that it is possible to reach all vertices of the graph from $r$ by moving along the directed edges.

You are given a directed tree with $N$ vertices (numbered $1$ through $N$). You may perform the following operation on it any number of times (including zero):

• Choose some edge which currently exists in the graph.
• Remove this edge.
• Add a new edge in such a way that the graph becomes a directed tree again.

What is the smallest number of operations which need to be performed in order to make the directed tree rooted?

### Input

• The first line of the input contains a single integer $T$ denoting the number of test cases. The description of $T$ test cases follows.
• The first line of each test case contains a single integer $N$.
• Each of the next $N-1$ lines contains two space-separated integers $u$ and $v$ denoting that there is a directed edge from $u$ to $v$ in the tree.

### Output

For each test case, print a single line containing one integer ― the smallest number of operations we need to perform to create a rooted directed tree.

### Constraints

• $1\le N\le {10}^{4}$
• $1\le u,v\le N$
• the graph described on the input is a directed tree
• the sum of $N$ over all test cases does not exceed ${10}^{5}$

Subtask #1 (20 points): for each edge, $u=1$ or $v=1$ (the graph is a directed star)

Subtask #2 (80 points): original constraints

2
3
1 2
3 2
3
1 2
2 3

1
0

### Explanation

Example case 1: We can delete the edge from vertex $3$ to vertex $2$ and insert an edge from $3$ to $1$. Then, the graph becomes a rooted directed tree with vertex $3$ as the root. However, there are many other possible solutions.

Example case 2: The graph is already a rooted directed tree; the root is vertex $1$.

solution->

According to this problem we have given a graph and we have to convert this graph to a rooted directed tree means every node should have only one incoming edge and 1 outgoing edge.

So my solution approach says that we will store the frequency of incoming edge of each vertex and

if I found incoming edge frequency greater then 1 so will delete this edge and add to the new node so my final answer will increase by ans+=freq[i]-1;

after doing the same for all vertex I will get my final answer.

my code is mentioned below. Please check it .

wait there is no need of creating any 2D boolean array sorry for that so delete that part of code.

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