Copil Copac is given a list of $n-1$ edges describing a tree of $n$ vertices. He decides to draw it using the following algorithm:

• Step $0$: Draws the first vertex (vertex $1$). Go to step $1$.
• Step $1$: For every edge in the input, in order: if the edge connects an already drawn vertex $u$ to an undrawn vertex $v$, he will draw the undrawn vertex $v$ and the edge. After checking every edge, go to step $2$.
• Step $2$: If all the vertices are drawn, terminate the algorithm. Else, go to step $1$.

The number of readings is defined as the number of times Copil Copac performs step $1$.

Find the number of readings needed by Copil Copac to draw the tree.

Input

Each test contains multiple test cases. The first line of input contains a single integer $t$ ($1\le t\le 10^4$) — the number of test cases. The description of test cases follows.

The first line of each test case contains a single integer $n$ ($2\le n\le 2\cdot 10^5$) — the number of vertices of the tree.

The following $n-1$ lines of each test case contain two integers $u_i$ and $v_i$ ($1\le u_i,v_i\le n$, $u_i\ne v_i$) — indicating that $(u_i,v_i)$ is the $i$-th edge in the list. It is guaranteed that the given edges form a tree.

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

Output

For each test case, output the number of readings Copil Copac needs to draw the tree.

Example

input
2
6
4 5
1 3
1 2
3 4
1 6
7
5 6
2 4
2 7
1 3
1 2
4 5
output
2
3

Note

In the first test case:

After the first reading, the tree will look like this:

After the second reading:

Therefore, Copil Copac needs $2$ readings to draw the tree.

解题思路

给所有的边从上到下依次进行编号(1~n),然后从顶点1开始建树，每次只能添加编号比当前编号大的边，如果出现编号小的边就相当于从新做了一次操作，操作数就加一

代码实现

#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
vector<pair<int,int>>s[N];
int n,ans;

void dfs(int v,int oj,int w=0){
	ans=max(ans,w);
	for(pair<int,int> i: s[v]){
		if(i.second!=oj)
		dfs(i.first,i.second,w+(i.second<oj));     //如果下一条边的编号小于当前边，操作数+1 
	}
}

void solve(){
	cin>>n;
	for(int i=1;i<=n;i++)s[i].clear();
	
	for(int i=1;i<n;i++){
		int a,b;
		cin>>a>>b;
		s[a].push_back({b,i});   //存边和边的编号 
		s[b].push_back({a,i});
	} 
	ans=0;
	dfs(1,0);
	ans++;
	cout<<ans<<endl;
}

int main(){
	int t;
	cin>>t;
	while(t--)
	solve();
	return 0;
}