#include "bits/stdc++.h"
using namespace std;
#define FAST ios_base::sync_with_stdio(false); cin.tie(0);
#define pb push_back
#define eb emplace_back
#define ins insert
#define f first
#define s second
#define cbr cerr<<"hi\n"
#define mmst(x, v) memset((x), v, sizeof ((x)))
#define siz(x) ll(x.size())
#define all(x) (x).begin(), (x).end()
#define lbd(x,y) (lower_bound(all(x),y)-x.begin())
#define ubd(x,y) (upper_bound(all(x),y)-x.begin())
mt19937 rng(chrono::steady_clock::now().time_since_epoch().count()); //can be used by calling rng() or shuffle(A, A+n, rng)
inline long long rand(long long x, long long y) { return rng() % (y+1-x) + x; } //inclusivesss
string inline to_string(char c) {string s(1,c);return s;} template<typename T> inline T gcd(T a,T b){ return a==0?llabs(b):gcd(b%a,a); }
using ll=long long;
using ld=long double;
#define FOR(i,s,e) for(ll i=s;i<=ll(e);++i)
#define DEC(i,s,e) for(ll i=s;i>=ll(e);--i)
using pi=pair<ll,ll>; using spi=pair<ll,pi>; using dpi=pair<pi,pi>;
long long LLINF = 1e18;
int INF = 1e9+1e6;
#define MAXN (100006)
ll n;
vector<int> v[MAXN];
pair<ll, vector<ll>> solve1() {
ll ans = 0;
vector<ll> A(n+1, -1), at(n+1, 0);
FOR(i,1,n)at[i]=i;
function<void(ll,ll)>dfs=[&](ll x,ll p) {
for(auto i:v[x]) if(i^p) {
dfs(i, x);
if(at[i] == i) {
swap(at[i], at[x]), ans += 2;
}
}
};
dfs(1,1);
if(at[1]==1) for(auto i:v[1]) { ans+=2, swap(at[1],at[i]); break; }
FOR(i,1,n)A[at[i]]=i;
A.erase(A.begin());
return {ans, A};
}
ll re;
bitset<MAXN> done;
spi diam() {
ll r = 1, mx = 0;
FOR(i,1,n) if(!done[i]) r=i;
function<void(ll,ll,ll)>dfs=[&](ll x,ll p,ll d){
if(d>mx)mx=d,r=x;
for(auto i:v[x]) if(i^p && !done[i]) dfs(i,x,d+1);
};
dfs(r,r,0);
ll r1=r; mx = 0;
dfs(r,r,0);
return {mx,{r1,r}};
}
pair<ll, vector<ll>> solve2() {
ll ans = 0;
vector<ll> A(n+1);
FOR(i,1,n) A[i] = i;
while(n - re) {
if(n - re == 3) {
ans += 4;
vector<ll> lol;
FOR(i,1,n) if(!done[i]) lol.eb(i);
FOR(i,0,2) A[lol[i]]=lol[(i+1)%3];
break;
}
spi x = diam();
done[x.s.f] = done[x.s.s] = 1;
re += 2;
swap(A[x.s.f], A[x.s.s]);
ans += x.f * 2;
}
A.erase(A.begin());
return {ans, A};
}
int main() {
FAST
cin>>n;
FOR(i,2,n){
ll a,b;
cin>>a>>b;
v[a].eb(b), v[b].eb(a);
}
auto mi = solve1();
auto mx = solve2();
cout<<mi.f<<' '<<mx.f<<'\n';
for(auto i:mi.s) cout<<i<<' '; cout<<'\n';
for(auto i:mx.s) cout<<i<<' '; cout<<'\n';
}
