#include <algorithm>
#include <iostream>
#include <iomanip>
#include <cstring>
#include <string>
#include <vector>
#include <cstdio>
#include <cmath>
#include <map>
using namespace std;

const int N = 100000;
const int M = 18;

int n;
vector <int> son[N+1];
vector <int> e[N+1];
int F[N+1][M+1];
bool done[N+1][M+1];
int G[N+1];
int G_which[N+1];

void dfs(int cur, int from) {
	for(int i = 0; i < e[cur].size(); i++) {
		if(e[cur][i] == from) continue;
		son[cur].push_back(e[cur][i]);
		dfs(e[cur][i], cur);
	}
}

int f(int cur, int need);
int g(int cur);

// the answer for subtree rooted at cur
int g(int cur) {
	if(G[cur] != -1)
		return G[cur];
	int maxv = -1, which = 0, cnt = 0;
	for(int i = 0; i < son[cur].size(); i++) {
		int v = g(son[cur][i]);
		if(v > maxv) {
			maxv = v;
			which = son[cur][i];
			cnt = 0;
		}
		if(v == maxv)
			cnt ++;
	}
	if(maxv == -1)
		G[cur] = 0;
	else if(cnt > 1) {
		G[cur] = maxv + 1;
	} else {
		if(f(which, maxv) == 0)
			G[cur] = maxv + 1;
		else
			G[cur] = maxv;
	}
	return G[cur];
}

// what is the max number k such that
// if we append a chain with k nodes at cur as parent, the answer is at most need
// if even k = 0 is not ok, return -1
int f(int cur, int need) {
	if(G[cur] != -1) {
		if(need < G[cur]) return -1;
		if(need > G[cur] + 17) return (1<<22);
	}
	if(done[cur][need - G[cur]])
		return F[cur][need - G[cur]];
	
	done[cur][need - G[cur]] = true;
	int ret = -1;
	if(son[cur].size() == 0) {
		ret = (need == 0 ? 0 : (1 << (min(need, 22) - 1)));
	} else {
		int which = 0, val = -1;
		for(int i = 0; i < son[cur].size(); i++) {
			int v = g(son[cur][i]) + 1;
			if(v > val) {
				val = v;
				which = i;
			}
		}
		int otherMax = 0;
		int nextF = -1;
		for(int i = 0; i < son[cur].size(); i++) {
			if(i == which) {
				nextF = f(son[cur][i], need) - 1;
			} else {
				otherMax = max(otherMax, g(son[cur][i]) + 1);
			}
		}
		if(nextF >= 0 && otherMax <= need) {
			int lef = need - otherMax;
			ret = min(nextF, lef == 0 ? 0 : (1 << (min(lef, 22) - 1)));
		}
	}
	F[cur][need - G[cur]] = ret;
	return ret;
}

int MAIN()
{
	int T;
	scanf("%d", &T);
	while(T--) {
		scanf("%d", &n);
		for(int i = 0; i <= n; i++) {
			e[i].clear();
			son[i].clear();
		}
		for(int i = 0; i < n-1; i++) {
			int a, b;
			scanf("%d %d", &a, &b);
			
			e[a].push_back(b);
			e[b].push_back(a);
		}
		memset(done, false, sizeof(done));
		memset(G, 0xff, sizeof(G));
		dfs(1, -1);
		printf("%d\n", g(1));
	}
	return 0;
}

int main()
{
	int start = clock();
	#ifdef LOCAL_TEST
		freopen("100000_1.in", "r", stdin);
		freopen("100000_1.out", "w", stdout);
	#endif
	int ret = MAIN();
	return ret;
}