Problem:
Score : 900 points

Problem Statement

We have a tree with N vertices.
The vertices are numbered 0 through N - 1, and the i-th edge (0 ≤ i < N - 1) comnnects Vertex a_i and b_i.
For each pair of vertices u and v (0 ≤ u, v < N), we define the distance d(u, v) as the number of edges in the path u-v.

It is expected that one of the vertices will be invaded by aliens from outer space.
Snuke wants to immediately identify that vertex when the invasion happens.
To do so, he has decided to install an antenna on some vertices.

First, he decides the number of antennas, K (1 ≤ K ≤ N).
Then, he chooses K different vertices, x_0, x_1, ..., x_{K - 1}, on which he installs Antenna 0, 1, ..., K - 1, respectively.
If Vertex v is invaded by aliens, Antenna k (0 ≤ k < K) will output the distance d(x_k, v).
Based on these K outputs, Snuke will identify the vertex that is invaded.
Thus, in order to identify the invaded vertex no matter which one is invaded, the following condition must hold:

For each vertex u (0 ≤ u < N), consider the vector (d(x_0, u), ..., d(x_{K - 1}, u)). These N vectors are distinct.

Find the minumum value of K, the number of antennas, when the condition is satisfied.

Constraints

2 ≤ N ≤ 10^5

0 ≤ a_i, b_i < N

The given graph is a tree.

Input

Input is given from Standard Input in the following format:

N
a_0 b_0
a_1 b_1
:
a_{N - 2} b_{N - 2}

Output

Print the minumum value of K, the number of antennas, when the condition is satisfied.

Sample Input 1

5
0 1
0 2
0 3
3 4

Sample Output 1

2

For example, install an antenna on Vertex 1 and 3.
Then, the following five vectors are distinct:

(d(1, 0), d(3, 0)) = (1, 1)

(d(1, 1), d(3, 1)) = (0, 2)

(d(1, 2), d(3, 2)) = (2, 2)

(d(1, 3), d(3, 3)) = (2, 0)

(d(1, 4), d(3, 4)) = (3, 1)

Sample Input 2

2
0 1

Sample Output 2

1

For example, install an antenna on Vertex 0.

Sample Input 3

10
2 8
6 0
4 1
7 6
2 3
8 6
6 9
2 4
5 8

Sample Output 3

3

For example, install an antenna on Vertex 0, 4, 9.

Solution in C++14 (Clang 3.8.0) (judge verdict: Wrong Answer)

#define  _CRT_SECURE_NO_WARNINGS
#include <stdio.h>
#include <algorithm>
#include <utility>
#include <functional>
#include <cstring>
#include <queue>
#include <stack>
#include <math.h>
#include <iterator>
#include <vector>
#include <string>
#include <set>
#include <math.h>
#include <iostream>
#include<map>
#include <iomanip>
#include <time.h>
#include <stdlib.h>
#include <list>
#include <typeinfo>
#include <list>
#include <set>
#include <cassert>
#include<fstream>
#include <cstdlib>
using namespace std;
#define Ma_PI 3.141592653589793
#define eps 0.00000001
#define LONG_INF 3000000000000000000
#define GOLD 1.61803398874989484820458
#define MAX_MOD 1000000007
#define REP(i,n) for(long long i = 0;i < n;++i)
#define seg_size 524288
#define PI 3.1415926535
using namespace std;
vector<int> vertexs[200000];
long long dp[2][200000] = {};
long long solve(long long now, long long back) {
	long long na = 0;
	long long mew = 0;
	for (int i = 0;i < vertexs[now].size();++i) {
		if (vertexs[now][i] != back) {
			solve(vertexs[now][i],now);
			na += dp[1][vertexs[now][i]];
			mew = max(mew, dp[1][vertexs[now][i]]);
		}
	}
	dp[0][now] = na - mew;
	dp[1][now] = na - mew + 1;
	return 0;
}
int main() {
	int n;
	cin >> n;
	REP(i, n-1) {
		int a, b;
		cin >> a >> b;
		vertexs[a].push_back(b);
		vertexs[b].push_back(a);
	}
	for (int i = 0;i < 200000;++i) {
		dp[0][i] = LONG_INF;
		dp[1][i] = LONG_INF;
	}
	int now_itr = 0;
	int maximum = 0;
	for (int i = 0;i < n;++i) {
		if (maximum < vertexs[i].size()) {
			maximum = vertexs[i].size();
			now_itr = i;
		}
	}
	solve(now_itr, -1);
	cout << max(1LL,dp[0][now_itr]) << endl;
	return 0;
}