/*
Baltic 15-net (accepted, looked at solution)
Key ideas:
- Simply link the ith leaf node to the i+(# of leaves)/2 node.
- Account for the case where (# of leaves) is odd.
*/
#include<iostream>
#include<vector>
using namespace std;
vector<int>s;
vector<vector<int>>t;//1 IDX
bool f[500005]={};
int N;
void dfs(int u){
if(f[u])return;
f[u]=1;
if(t[u].size()==1)s.push_back(u);
for(int&i:t[u])dfs(i);
}
int main(){
cin>>N;
for(int i=0;i<=N+5;++i)t.push_back(vector<int>());
for(int i=0;i<N-1;++i){
int a,b;cin>>a>>b;
t[a].push_back(b);
t[b].push_back(a);
}
dfs(1);
int k=s.size();
if(k%2){
cout<<k/2+1<<'\n';
for(int i=0;i<k/2;++i){
cout<<s[i]<<' '<<s[i+k/2+1]<<'\n';
}
cout<<s[k/2]<<' '<<1<<'\n';
}else{
cout<<k/2<<'\n';
for(int i=0;i<k/2;++i){
cout<<s[i]<<' '<<s[i+k/2]<<'\n';
}
}
// free memory
vector<vector<int>>().swap(t);
vector<int>().swap(s);
return 0;
}
