#include <bits/stdc++.h>

 using namespace std;

 #define rep(i,n) for (int i = 0; i < (n); ++i)

 #define For(i,s,t) for (int i = (s); i <= (t); ++i)

 #define rFor(i,t,s) for (int i = (t); i >= (s); --i)

 #define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)

 #define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)

 #define pr(x) cout << #x << " = " << x << "  "

 #define prln(x) cout << #x << " = " << x << endl

 #define LOWBIT(x) ((x)&(-x))

 #define ALL(x) x.begin(),x.end()

 #define INS(x) inserter(x,x.begin())

 #define ms0(a) memset(a,0,sizeof(a))

 #define msI(a) memset(a,inf,sizeof(a))

 #define msM(a) memset(a,-1,sizeof(a))

 #define pii pair<int,int>

 #define piii pair<pair<int,int>,int>

 #define mp make_pair

 #define pb push_back

 #define fi first

 #define se second

 inline int gc(){

     static const int BUF = 1e7;

     static char buf[BUF], *bg = buf + BUF, *ed = bg;

     if(bg == ed) fread(bg = buf, , BUF, stdin);

     return *bg++;

 } 

 inline int ri(){

     int x = , f = , c = gc();

     for(; c<||c>; f = c=='-'?-:f, c=gc());

     for(; c>&&c<; x = x* + c - , c=gc());

     return x*f;

 }

 typedef long long LL;

 typedef unsigned long long uLL;

 const int inf = 1e9 + ;

 const LL mod = 1e9 + ;

 const int maxN = 2e5 + ;

 struct Node{

     vector< pii > next;

 };

 int n, ans;

 // dp[i]表示以i号城市为首都所要反转的道路数

 int dp[maxN];

 int vis[maxN];

 Node nodes[maxN];

 queue< int > Q;

 // 求dp[x]

 inline int dfs(int x) {

     if(nodes[x].next.empty()) return ;

     int ret = ;

     foreach(i, nodes[x].next) {

         int y = i->fi, z = i->se;

         if(vis[y]) continue;

         vis[x] = ;

         if(z == -) ++ret;

         ret += dfs(y);

     }

     return ret;

 }

 // 如果dp[i]已知，假设j号城市与i号城市相连，由于整个城市图是树形结构，

 // 因此dp[j]与dp[i]除了他们的连线有分歧，其余应该反转的道路数量都是一样的

 // 因此可由已算出来的1号城市为中心，向外做BFS

 inline void bfs() {

     ms0(vis);

     vis[] = ;

     Q.push();

     ans = dp[];

     while(!Q.empty()) {

         int tmp = Q.front();

         Q.pop();

         foreach(i, nodes[tmp].next) {

             int y = i->fi, z = i->se;

             if(vis[y]) continue;

             vis[y] = ;

             dp[y] = dp[tmp] + z;

             ans = min(ans, dp[y]);

             Q.push(y);

         }

     }

 }

 int main(){

     while(cin >> n) {

         ms0(vis);

         rep(i, n + ) nodes[i].next.clear();

         rep(i, n-) {

             int x, y;

             cin >> x >> y;

             nodes[x].next.push_back(mp(y, ));

             nodes[y].next.push_back(mp(x, -));

         }

         dp[] = dfs();

         bfs();

         cout << ans << endl;

         For(i, , n) if(ans == dp[i]) cout << i << " ";

         cout << endl;

     }

     return ;

 }