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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<bits/stdc++.h>
using namespace std;

#define endl '\n'
#define lfs cout<<fixed<<setprecision(10)
#define ALL(a)  (a).begin(),(a).end()
#define ALLR(a)  (a).rbegin(),(a).rend()
#define spa << " " <<
#define test cout<<"test"<<endl;
#define fi first
#define se second
#define MP make_pair
#define PB push_back
#define EB emplace_back
#define rep(i,n,m) for(ll i = (n); i < (ll)(m); i++)
#define rrep(i,n,m) for(ll i = (m) - 1; i >= (ll)(n); i--)
using ll = long long;
using ld = long double;
const ll MOD = 1e9+7;
//const ll MOD = 998244353;
const ll INF = 1e18;
using P = pair<ll, ll>;
template<typename T>
void chmin(T &a,T b){if(a>b)a=b;}
template<typename T>
void chmax(T &a,T b){if(a<b)a=b;}
void pmod(ll &a,ll b){a=(a+b)%MOD;}
void pmod(ll &a,ll b,ll c){a=(b+c)%MOD;}
void qmod(ll &a,ll b){a=(a*b)%MOD;}
void qmod(ll &a,ll b,ll c){a=(b*c)%MOD;}
ll median(ll a,ll b, ll c){return a+b+c-max({a,b,c})-min({a,b,c});}
void ans1(bool x){if(x) cout<<"Yes"<<endl;else cout<<"No"<<endl;}
void ans2(bool x){if(x) cout<<"YES"<<endl;else cout<<"NO"<<endl;}
void ans3(bool x){if(x) cout<<"Yay!"<<endl;else cout<<":("<<endl;}
template<typename T1,typename T2>
void ans(bool x,T1 y,T2 z){if(x)cout<<y<<endl;else cout<<z<<endl;}  
template<typename T>
void debug(vector<vector<T>>&v,ll h,ll w){for(ll i=0;i<h;i++)
{cout<<v[i][0];for(ll j=1;j<w;j++)cout spa v[i][j];cout<<endl;}};
void debug(vector<string>&v,ll h,ll w){for(ll i=0;i<h;i++)
{for(ll j=0;j<w;j++)cout<<v[i][j];cout<<endl;}};
template<typename T>
void debug(vector<T>&v,ll n){if(n!=0)cout<<v[0];
for(ll i=1;i<n;i++)cout spa v[i];cout<<endl;};
template<typename T>
vector<vector<T>>vec(ll x, ll y, T w){
  vector<vector<T>>v(x,vector<T>(y,w));return v;}
ll gcd(ll x,ll y){ll r;while(y!=0&&(r=x%y)!=0){x=y;y=r;}return y==0?x:y;}
vector<ll>dx={1,0,-1,0,1,1,-1,-1};
vector<ll>dy={0,1,0,-1,1,-1,1,-1};
template<typename T>
vector<T> make_v(size_t a,T b){return vector<T>(a,b);}
template<typename... Ts>
auto make_v(size_t a,Ts... ts){
  return vector<decltype(make_v(ts...))>(a,make_v(ts...));
}
ostream &operator<<(ostream &os, pair<ll, ll>&p){
  return os << p.first << " " << p.second;
}  
ll MAX_V = 2e5 + 5;
vector<ll>t(MAX_V);
pair<ll,bool> dfs(ll k,vector<vector<ll>>&g){
  t[k]=1;
  ll ret=0;
  ll cnt=0,cnt2=0,cnt3=0;
  for(ll i=0;i<g[k].size();i++){
    if(t[g[k][i]]==0){
      auto p=dfs(g[k][i],g); 
      if(p.fi==0)cnt++;
      if(p.fi>=1) cnt2++;
      if(p.se)cnt3++;
      ret+=p.fi;
    }
  }
  //cout<<k spa ret spa cnt<<endl;
  if(k==0){
    //cout<<cnt spa cnt2 spa cnt3 spa ret<<endl;
    if(cnt3!=cnt2&&cnt2==1&&cnt<=1)ret++;
    ret+=max(0LL,cnt-1);
    return MP(ret,false);
  }
  bool sw=false;
  if(cnt==0&&(cnt3==cnt2||cnt2>=2))sw=true;
  ret+=max(0LL,cnt-1);
  return MP(ret,sw);
}
int main(){
  cin.tie(nullptr);
  ios_base::sync_with_stdio(false);
  ll res=0,buf=0;
  bool judge = true;
  ll n;cin>>n;
  vector<vector<ll>>g(n);
  rep(i,0,n-1){
    ll a,b;cin>>a>>b;
    g[a].PB(b);
    g[b].PB(a);
  }
  auto p=dfs(0,g);
  cout<<max(1LL,p.fi)<<endl;
  return 0;
}