// Success consists of going from failure to failure without loss of enthusiasm
#include <bits/stdc++.h>
using namespace std;
#define nl '\n'
#define pb push_back
#define pf push_front
#define mp make_pair
#define f first
#define s second
#define sz(x) int(x.size())
template<class T> using V = vector<T>;
using pi = pair<int, int>;
using vi = V<int>;
using vpi = V<pi>;
using ll = long long;
using pl = pair<ll, ll>;
using vpl = V<pl>;
using vl = V<ll>;
using db = long double;
template<class T> using pq = priority_queue<T, V<T>, greater<T>>;
const int MOD = 1e9 + 7;
const ll INFL = ll(1e17);
const int LG = 18;
int main() {
cin.tie(0)->sync_with_stdio(0);
int N; cin >> N;
vi deg(N);
V<vi> adj(N); for(int i = 0; i < N - 1; i++) {
int u, v; cin >> u >> v; --u, --v;
adj[u].pb(v);
adj[v].pb(u);
deg[u]++, deg[v]++;
}
int R = -1; for(int i = 0; i < N; i++) if (deg[i] > 1) R = i;
vi L;
function<void(int, int)> dfs = [&](int u, int p) {
bool leaf = 1;
for(auto& v : adj[u]) if (v != p) {
leaf = 0;
dfs(v, u);
}
if (leaf) L.pb(u);
};
dfs(R, -1);
int K = (sz(L) + 1) / 2;
cout << K << nl;
for(int i = 0; i < K; i++) {
cout << L[i] + 1 << " " << L[i + sz(L) / 2] + 1 << nl;
}
exit(0-0);
}
